Brewster’s angle

In parallel polarization the incident angle at which there is no reflection is called Brewster’s angle.

\rho _{parallel} = 0.

As   \rho _{parallel} = \frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}.

\frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}=0.

(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})=0.

\eta _{2}\cos \theta _{t}=\eta _{1}\cos \theta _{i}.

by squaring on both sides  \eta^{2} _{2}\cos^{2} \theta _{t}=\eta^{2} _{1}\cos^{2} \theta _{i}.

\eta^{2}_{2}(1-\sin^{2} \theta _{t})=\eta^{2} _{1}(1-\sin^{2} \theta _{i})....EQN(I)

By using Snell’s law  \frac{\sin \theta _{i}}{\sin \theta _{t}} =\sqrt{\frac{\mu _{2}\epsilon _{2}}{\mu _{1}\epsilon _{1}}} =\frac{r_{2}}{r_{1}}.

using the above equation  \sin^{2} \theta _{t} =\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i}. substituting this in EQN (I).

\eta^{2}_{2}(1-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=\eta^{2} _{1}(1-\sin^{2} \theta _{i}).

(\eta^{2}_{2}-\eta^{2}_{2}\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=(\eta^{2} _{1}- \eta^{2} _{1}\sin^{2} \theta _{i}).

after simplification    \sin^{2} \theta _{i}(\eta^{2} _{1}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\eta^{2} _{2} )=(\eta^{2} _{1}-\eta^{2} _{2 }).

as \eta _{1} = \sqrt{\frac{\mu _{1}}{\epsilon _{1}}}  and  \eta _{2} = \sqrt{\frac{\mu _{2}}{\epsilon _{2}}}.

\sin^{2} \theta _{i}(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\frac{\mu _{2}}{\epsilon _{2}} )=(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{2}}{\epsilon _{2}})..

by simplification  \sin^{2} \theta _{i}= \frac{(1-\frac{\mu _{2}\epsilon _{1}}{\mu _{1}\epsilon _{2}} )}{(1-\frac{\epsilon^{2} _{1}}{\epsilon^{2} _{2}})}.

here \theta _{i} is called as Brewster’s angle.

Let us assume two mediums are lossless dielectrics and are non-magnetic  then

\mu _{1} =\mu _{2}=\mu _{0}.

\sin^{2} \theta _{Brewster}= \frac{1}{(1+\frac{\epsilon _{1}}{\epsilon _{2}})}.

\sin \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{2}+\epsilon _{1}}}.

\tan \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{1}}} = \frac{r_{2}}{r_{1}}.

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Parallel Polarization

Parallel Polarizaton:-

Parallel polarization means \overrightarrow{E} field lies in the XZ-plane (y=0) that is the plane of incidence, the figure illustrates the case of parallel polarization

then the incident and reflected fields is given by

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(\overrightarrow{k}.\overrightarrow{r})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}}).(k_{i}\cos \theta _{i}\overrightarrow{a_{z}}+k_{i}\sin \theta _{i}\overrightarrow{a_{x}})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(k_{i}\cos \theta _{i}z+xk_{i}\sin \theta _{i})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j\beta _{1}(z\cos \theta _{i}+x\sin \theta _{i})}

\overrightarrow{H_{is}} = \frac{E_{i}}{\eta _{1}}e^{-j\beta _{1}(z\cos \theta _{i}+x\sin \theta _{i})}\overrightarrow{a_{y}}

now the reflected wave is 

\overrightarrow{E_{rs}} = E_{r}(\sin \theta_{r }\overrightarrow{a_{z}}+ \cos \theta_{r }\overrightarrow{a_{x}} )e^{-j\beta _{1}(-z\cos \theta _{r}+x\sin \theta _{r})}

\overrightarrow{H_{rs}} =\frac{ H_{r}}{\eta _{1}}e^{-j\beta _{1}(-z\cos \theta _{r}+x\sin \theta _{r})}\overrightarrow{a_{y}}

since k_{i}=k_{r}=\beta _{1} =\omega \sqrt{\mu_{1} \epsilon_{1} }

let’s find out  \overrightarrow{k}  then by using the equation \overrightarrow{k}.\overrightarrow{E_{s}} =0\overrightarrow{H_{s}} = \frac{k}{\omega \mu }X \overrightarrow{E_{s}} = \overrightarrow{a_{k}}X \frac{E_{s}}{\eta } .

and the transmitted fields in the second  medium is  

\overrightarrow{E_{ts}} = E_{t}(-\sin \theta_{t }\overrightarrow{a_{z}}+ \cos \theta_{t}\overrightarrow{a_{x}} )e^{-j\beta _{2}(z\cos \theta _{t}+x\sin \theta _{t})}

\overrightarrow{H_{ts}} = \frac{H_{t}}{\eta _{2}}e^{-j\beta _{2}(z\cos \theta _{t}+x\sin \theta _{t})}\overrightarrow{a_{y}}

where \beta _{2} = \omega \sqrt{\mu _{2}\epsilon _{2}}.

Transmission coefficient:-

as \theta _{r} = \theta _{i} and also that the tangential components of \overrightarrow{E} and \overrightarrow{H} are continuous at the boundary z=0.


(E_{i}+E_{r}) \cos \theta _{i} = E_{t}\cos \theta _{t}

H_{tan1}-H_{tan2} = \overrightarrow{J_{s}} , H_{tan1} = H_{tan2}  since \overrightarrow{J_{s}} =0

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}=\frac{E_{t}}{\eta _{2}}------EQN(1)

E_{i}+E_{r}=E_{t}\frac{\cos \theta _{t}}{\cos \theta _{i}}------EQN(2)

(1)+(2) implies 

2E_{i}=E_{t}(\frac{\eta _{1}}{\eta _{2}}+\frac{\cos \theta _{t}}{\cos \theta _{i}})

2E_{i}=E_{t}\frac{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}{\eta _{2}\cos \theta _{i}}

\tau _{parallel} = \frac{E_{t}}{E_{i}} = \frac{2\eta _{2}\cos \theta _{i}}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}.

Reflection coefficient:-

from EQN (1)

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}=\frac{E_{t}}{\eta _{2}}------EQN(1)

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}= \frac{2E_{i}\eta _{2}\cos \theta _{i}}{\eta _{2}(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}

after simplification

E_{r}\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}=E_{i}

\rho _{parallel} = \frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}

1+\rho _{parallel} = \tau _{parallel}(\frac{\cos \theta _{t}}{\cos \theta _{i}}).

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Surface impedance

At high frequencies, the current is almost confined to a very thin sheet at the surface of the conductor which is used in many applications.

The  surface impedance may be defined as the ratio of the tangential component of the electric field \overrightarrow{E_{tan}} at the surface of the conductor to the current density (linear) \overrightarrow{J_{s}} which flows due to this electric field.

given as Z_{s} (or) \eta _{s} = \frac{\overrightarrow{E_{tan}}}{\overrightarrow{J_{s}}}.

\overrightarrow{E_{tan}}   is the Electric field strength parallel to and at the surface of the conductor.

and \overrightarrow{J}  is the total linear current density which flows due to \overrightarrow{E_{tan}}.

The \overrightarrow{J_{s}} represents the total conduction per meter width flowing in this sheet.

Let us consider a conductor of the type plate, is placed at the surface y=0 and the current distribution in the y-direction is given by 


Assume that the depth of penetration (\delta) is very much less compared with the thickness of the conductor.

J_{s}= \int_{0}^{\infty } \overrightarrow{J}.\overrightarrow{dy}

J_{s}= \int_{0}^{\infty } J_{o}e^{-\gamma y}dy

J_{s}= J_{o}(e^{-\gamma y})_{0}^{\infty }

J_{s}= \frac{J_{o}}{\gamma }

from ohm’s law \overrightarrow{J_{o}} = \sigma \overrightarrow{E_{tan}}  

E = \frac{J_{o}}{\sigma } .

then  \eta _{s} = \frac{\gamma }{\sigma } .

Z_{s}  (or)  \eta _{s} = \frac{\gamma }{\sigma } .

we know that \gamma = \sqrt{j\omega \mu (\sigma +j\omega \epsilon )}  

for good conductors \sigma > > \omega \epsilon .

then \gamma \approx \sqrt{j\omega \mu \sigma } 

\eta _{s} = \frac{\gamma }{\sigma } = \sqrt{\frac{j\omega \mu }{\sigma }} .

therefore the surface impedance of a plane good conductor which is very much thicker than the skin depth is equal to the characteristic impedance of the conductor.

This impedance is also known s input impedance of the conductor when viewed as transmission line conducting energy into the interior of metal.

when the thickness of the plane conductor is not greater compared to the depth of penetration , reflection of wave occurs at the back surface of the conductor.




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oblique incidence

when a uniform plane wave  is incident obliquely (making an angle \theta _{i} other than 90^{o}) to the boundary between the two media then it is known as oblique incidence.

Now consider the situation that is more general case  that is the oblique incidence.

In this case the EM wave (incident wave) not strikes normally the boundary. i.e,  the incident wave is not propagating  along any standard axes (like x,y and z).

Therefore EM wave is moving in a random direction then the general form is    \overrightarrow{E} = E_{o} \cos (\omega t-\overrightarrow{k}.\overrightarrow{r})  

it is also in the form \overrightarrow{E} = E_{o} \cos (\overrightarrow{k}.\overrightarrow{r}-\omega t).

then \overrightarrow{k} = k_{x}\overrightarrow{a_{x}}+k_{y}\overrightarrow{a_{y}}+k_{z}\overrightarrow{a_{z}} is called the wave number vector (or) the propagation vector.

and \overrightarrow{r} = x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}} is called the position vector (from origin to any point on the plane of incidence) , then the magnitude of \overrightarrow{k} is related to \omega according to the dispersion.

k^{2} = k_{x}^{2}+k_{y}^{2}+k_{z}^{2} = \omega ^{2}\mu \epsilon

\overrightarrow{k} X \overrightarrow{E} = \omega \mu \overrightarrow{H}.

\overrightarrow{k} X \overrightarrow{H} = -\omega \epsilon \overrightarrow{E}.

\overrightarrow{k} . \overrightarrow{H} =0.

\overrightarrow{k} . \overrightarrow{E} =0.

i.    \overrightarrow{E},\overrightarrow{H} and \overrightarrow{k} are mutually orthogonal.

ii.  \overrightarrow{E}  and  \overrightarrow{H} lie on the plane \overrightarrow{k} . \overrightarrow{r} = k_{x}x+k_{y}y+k_{z}z=constant.

then the \overrightarrow{H} field corresponding to \overrightarrow{E} field is \overrightarrow{H} = \frac{1}{\omega \mu } (\overrightarrow{k} X \overrightarrow{E}) = \frac{\overrightarrow{a_{k}} X \overrightarrow{E}}{\eta }.

Now choose oblique incidence of a uniform plane wave at a plane boundary.

the plane defined by the propagation vector \overrightarrow{k} and a unit normal vector \overrightarrow{a_{n}} to the boundary is called the plane of incidence.

the angle \theta _{i} between \overrightarrow{k} and \overrightarrow{a_{n}} is the angle of incidence.

both the incident and reflected waves are in medium 1 while the transmitted wave is in medium 2 .


\overrightarrow{E_{i}} =E_{i}\cos (k_{ix}x+k_{iy}y+k_{iz}z-\omega _{i}t)

\overrightarrow{E_{r}} =E_{i}\cos (k_{rx}x+k_{ry}y+k_{rz}z-\omega _{r}t)

\overrightarrow{E_{t}} =E_{t}\cos (k_{tx}x+k_{ty}y+k_{tz}z-\omega _{t}t).

the wave propagates 

  1.     \omega _{i}=\omega _{r}=\omega _{t}=\omega.
  2.     k_{ix} = k_{rx}=k_{tx}=k_{x}.
  3.    k_{iy} = k_{ry}=k_{ty}=k_{y}.

(1) indicates that all waves are propagating with same frequency. (2) and (3) shows that the tangential components of propagation vectors be continuous.

k_{i} \sin \theta _{i}=k_{r} \sin \theta _{r}  implies  k_{i} = k_{r} =\beta _{1} =\omega \sqrt{\mu _{1}\epsilon _{1}}    since \theta _{r}=\theta _{i}.

k_{i} \sin \theta _{i}=k_{t} \sin \theta _{t}   implies k_{t} =\beta _{2} =\omega \sqrt{\mu _{2}\epsilon _{2}}.

\frac{\sin \theta _{t}}{sin \theta _{i}}=\sqrt{\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}}

now velocity u=\frac{\omega }{k}.

then from Snell’s law       r_{1}\sin \theta _{i} = r_{2}\sin \theta _{t},  where r_{1} and r_{2}  are the refractive indices of the two media.




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Poynting theorem


Poynting theorem is used to get an expression for propagation of energy in a medium.

It gives the relation between energy stored in a time-varying magnetic field and the energy stored in time-varying electric field and the instantaneous power flow out of a given region.

EM waves propagates through space from source to destination. In order to find out power in a uniform plane wave it is necessary to develop a power theorem for the EM field known as poynting theorem.

The direction of power flow is perpendicular to \vec{E} and \overrightarrow{H} in the direction of plane containing \overrightarrow{E} and \overrightarrow{H}.

i.e, it gives the direction of propagation .

\overrightarrow{P} = \overrightarrow{E}X\overrightarrow{H}  Watts/m2  (or)   VA/m2.


from Maxwell’s  equations \overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\overrightarrow{J_{d}}

\overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{J}=\overrightarrow{\bigtriangledown } X \overrightarrow{H}-\frac{\partial \overrightarrow{D}}{\partial t}

the above equation has units of the form current density Amp/m2. When it gets multiplied by \overrightarrow{E} V/m. The total units  will  have of the form power per unit volume.

\overrightarrow{J}\rightarrow Amp/m2\overrightarrow{E}\rightarrow Volts/m.

EJ\rightarrow Amp. Volt/m3 \rightarrow Watts/m3 \rightarrow Power/volume.

\overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{E}.(\overrightarrow{\bigtriangledown } X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{E}.(\overrightarrow{\bigtriangledown } X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}

by using the vector identity 

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{H}.(\overrightarrow{\bigtriangledown }X \overrightarrow{E} )-\overrightarrow{E}.(\overrightarrow{\bigtriangledown }X \overrightarrow{H} )


\overrightarrow{H}.(\overrightarrow{\bigtriangledown }X \overrightarrow{E} )-\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}.

from the equation of Maxwell’s \overrightarrow{\bigtriangledown } X \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} = -\frac{\partial (\mu \overrightarrow{ H})}{\partial t}

\overrightarrow{H}.(-\frac{\partial (\mu \overrightarrow{ H})}{\partial t} )-\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\overrightarrow{E}.\overrightarrow{J}-\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}-\overrightarrow{H}.(\frac{\partial (\mu \overrightarrow{ H})}{\partial t} )------EQN(I)

by using the vector identity  \frac{\partial (\overrightarrow{A}.\overrightarrow{B})}{\partial t}=\overrightarrow{A}.\frac{\partial \overrightarrow{B}}{\partial t}+\overrightarrow{B}.\frac{\partial \overrightarrow{A}}{\partial t}

if \overrightarrow{A} = \overrightarrow{B}    \Rightarrow \frac{\partial (\overrightarrow{A}.\overrightarrow{A})}{\partial t}=2 \overrightarrow{A}.\frac{\partial \overrightarrow{A}}{\partial t}

\overrightarrow{A}.\frac{\partial \overrightarrow{A}}{\partial t}=\frac{1}{2}\frac{\partial A^{2}}{\partial t}

from EQN(I)   ,  \overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\overrightarrow{E}.\sigma\overrightarrow{E }-\frac{1}{2}\epsilon\frac{\partial E^{2}}{\partial t}-\frac{1}{2}\mu \frac{\partial H^{2}}{\partial t}

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\sigma E^{2}-\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})

by integrating the above equation by over  a volume 

\int_{v}\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) dv=-\int_{v}\sigma E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})dv

by converting the volume integral to surface integral

\oint_{s}(\overrightarrow{E} X \overrightarrow{H}). \overrightarrow{ds}=-\int_{v}\sigma E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})dv.

the above equation gives the statement of Poynting theorem.

Poynting theorem:-  

It states that the net power flowing out of a given volume is equal to the time rate of decrease in the energy stored with in that volume V and the ohmic losses.






Inductance of a Co-axial cable, Solenoid & Toroid

Inductance of a Co-axial cable:-

Consider a coaxial cable with inner conductor having radius a and outer conductor of radius b and the current is flowing in the cable along z-axis  in which the cable is placed such that the axis of rotation of the cable co-incides with z-axis.

the length of the co-axial cable be d meters.

we know that the \overrightarrow{H_{\phi }} between the region a< \rho < b  is \overrightarrow{H_{\phi }}=\frac{I}{2\pi \rho }\overrightarrow{a_{\phi }}

and \overrightarrow{B_{\phi }}=\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }}   since \overrightarrow{B_{\phi }}=\mu _{o}\overrightarrow{H_{\phi }}

as L = \frac{\lambda }{I}, here the flux linkage \lambda =1\phi

where \phi-Total flux coming out of the surface

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

Since the magnetic flux will be radial plane extending from \rho =a   to   \rho =b and z=0 to z=d.

\overrightarrow{ds_{\phi }} = d\rho dz \overrightarrow{a_{\phi }}

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

\phi = \oint_{s}\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }} .d\rho dz \overrightarrow{a_{\phi }}

\phi = \int_{\rho =a}^{b}\int_{z =0}^{d}\frac{\mu _{o}I}{2\pi \rho } d\rho dz

\phi = \frac{\mu _{o}I}{2\pi } ln[\frac{b}{a}] d

L=\frac{\phi }{I} =\frac{\mu _{o}d}{2\pi } ln[\frac{b}{a}]  Henries.


Inductance of a Solenoid:-

Consider a Solenoid of N turns and  let the current flowing inside it is  ‘I’ Amperes. The length of the solenoid is ‘l’ meters and ‘A’ is its cross sectional area.

\phi – Total flux coming out of solenoid.

flux linkage \lambda = N\phi      \Rightarrow \lambda = NBA----------EQN(I)        \because \frac{\phi }{A} = B

L = \frac{\lambda }{I}, from the definition

As B is the Magnetic flux density given B= \frac{flux}{unit area} =\frac{\phi }{A}

from EQN(I) ,

\lambda = N\mu _{o}HA-------EQN(II)  because B = \mu _{o}H

The field strength H of a solenoid is H = \frac{NI}{l} A/m

EQN (II) becomes  \lambda = N \mu _{o}\frac{NI}{l}A

\lambda = \frac{N^{2} \mu _{o}IA}{l}

from the inductance definition L = \frac{\lambda }{I}

L= \frac{N^{2} \mu _{o}A}{l}   Henries.

Inductance of a Toroid:-

Consider a toroidal ring with N-turns and carrying current I.

let the radius of the toroid be ‘R’ and the total flux emerging be \phi

then flux linkage  \lambda = N\phi

the magnetic flux density inside a toroid is given by B = \frac{\mu _{o}NI}{2\pi R}

\lambda = NBA

where A is the cross sectional area of the toroid then \lambda = N \frac{\mu_{o} NI}{2\pi R}A

\lambda = \frac{\mu_{o} N^{2}I}{2\pi R}A

L=\frac{\lambda }{I}

L=\frac{\mu_{o} N^{2}A}{2\pi R}  Henries.

if the toroid has a height ‘h’ , inner radius \rightarrow r_{1} and outer radius \rightarrow r_{2} then its Inductance is L=\frac{\mu_{o} N^{2}h}{2\pi }ln[\frac{r_{1}}{r_{2}}]   Henries.

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continuity equation

The continuity equation of the current is based on the principle of conservation of charge that is  the charge can neither be created not destroyed.

consider a closed surface S with a current density \overrightarrow{J}, then the total current I crossing the surface S is given by the volume V

The current coming out of the closed surface is I_{out} = \oint_{s} \overrightarrow{J}.\overrightarrow{ds}

since the direction of current is in the direction of positive charges, positive charges also move out of the surface because of the current I.

According to principle of conversation of charge, there must be decrease of an equal amount of positive charge inside the closed surface.

therefore the time rate of decrease of charge with in a given volume must be equal to the net outward current flow through the closed surface of the volume.

I_{out} = \frac{-dQ_{in}}{dt}= \oint_{s} \overrightarrow{J}.\overrightarrow{ds}

By Divergence theorem \oint_{s} \overrightarrow{J}.\overrightarrow{ds} = \oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv

\frac{-dQ_{in}}{dt}= \oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv

Q_{in}=\int_{v}\rho _{v}dv    implies  -\frac{dQ_{in}}{dt}=-\frac{d}{dt}\int_{v}\rho _{v}dv

for a constant surface the derivative becomes the partial derivative 

\oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv =-\int_{v}\frac{\partial \rho _{v}}{\partial t}dv   -this is for the whole volume.

for a differential volume \overrightarrow{\bigtriangledown }.\overrightarrow{J} dv =-\frac{\partial \rho _{v}}{\partial t}dv

\overrightarrow{\bigtriangledown }.\overrightarrow{J} =-\frac{\partial \rho _{v}}{\partial t} , which is called as continuity of current equation (or) Point form (or) differential form of the continuity equation.

This equation is derived based on the principle of conservation charge states that there can be no accumulation of charge at any point.

for steady  (dc) currents     \overrightarrow{\bigtriangledown }.\overrightarrow{J} =0 \Rightarrow -\frac{\partial \rho _{v}}{\partial t}=0

from \overrightarrow{\bigtriangledown }.\overrightarrow{J} =0. The total charge leaving a volume is the same as total charge entering it. Kirchhoff’s law follows this equation.

This continuity equation states that the current (or) the charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point.


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E due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density \rho _{L} C/m and lies on Z-axis from -\infty to +\infty.

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are (0,\rho ,0) ( a point on y-axis) and assume dQ is a small differential charge confirmed to a point  M (0,0,Z) as co-ordinates.

\therefore dQ produces a differential field \overrightarrow{dE} 

\overrightarrow{dE}=\frac{dQ}{4\pi \epsilon _{o}R^{2}}\widehat{a_{r}}

the position vector \overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }} and the corresponding unit vector \widehat{a_{r}} =\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{\rho ^{2}+Z^{2}}}

\therefore \overrightarrow{dE} =\frac{dQ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

therefore \overrightarrow{dE} =\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

then the Electric field strength \overrightarrow{E} produced by the infinite line charge distribution \rho _{L} is 

\overrightarrow{E} = \int \overrightarrow{dE}

\overrightarrow{E} = \int_{z=-\infty }^{\infty }\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

to solve this integral  let Z= \rho \tan \theta \Rightarrow dZ=\rho \sec ^{2}\theta d\theta

as Z\rightarrow -\infty \Rightarrow \theta \rightarrow \frac{-\pi }{2}

Z\rightarrow \infty \Rightarrow \theta \rightarrow \frac{\pi }{2}

\therefore \overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{(\rho ^{2}+\rho ^{2}\tan ^{2}\theta )^{\frac{3}{2}}})

\overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{\rho ^{3}\sec ^{3}\theta })

\overrightarrow{E} = \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }[\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} -\sin \theta \overrightarrow{a_{z} }d\theta +\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \cos\theta d\theta \overrightarrow{a_{\rho }}]

\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }[0+2\overrightarrow{a_{\rho }}]

\therefore \overrightarrow{E}= \frac{\rho _{L}}{2\pi\epsilon _{o}\rho }\overrightarrow{a_{\rho }} Newtons/Coulomb.

\overrightarrow{E} is a function of \rho   only, there is no \overrightarrow{a_{z}} component and \rho is the perpendicular distance from the point P to line charge distribution \rho _{L}.

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Energy Density in Electrostatic Fields

To determine the Energy present in an assembly of charges (or) group of charges one must first determine the amount of work necessary to assemble them.

It is seen that , when a unit positive charge is moved from infinity to a point in a field, the work is done by the external source and energy is expended.

If the external source is removed then the unit positive charge will be subjected to a force exerted by the field and will be moved in the direction of force.

Thus to hold the charge at a point in an electrostatic field, an external source has to do work , this energy gets stored in the form of Potential Energy when the test charge is hold at a point in a field.

when external source is removed , the Potential Energy gets converted to a Kinetic Energy.

In order to derive the expression for energy stored in electrostatics (i.e, the expression of such a Potential Energy)

Consider an empty space where there is no electric field at all, the Charge Q_{1} is moved from infinity to a point in the space ,let us say the point as P_{1}, this requires no work to be done to place a charge Q_{1} from infinity to a point P_{1} in empty space.

i.e, work done = 0 for placing a charge Q_{1} from infinity to a point P_{1} in empty space.

now another charge Q_{2} has to be placed from infinity to another point P_{2} . Now there has to do some work to place Q_{2} at P_{2} because there is an electric field , which is produced by the charge Q_{1} and Q_{2} is required to move against the field of Q_{1}.

Hence the work required to be done is  Potential=\frac{work done}{unit charge}

i.e, V = \frac{W}{Q} \Rightarrow W = V X Q .

\therefore Work done to position Q_{2} at P_{2} = V_{21} X Q_{2}.

Now the charge Q_{3} to be moved from infinity to P_{3} , there are electric fields due to Q_{1} and Q_{2}, Hence total work done is due to potential at P_{3} due to charge at P_{1} and Potential at P_{3} due to charge at P_{2}.

\therefore Work done to position Q_{3} at P_{3} = V_{31}Q_{3}+V_{32}Q_{3}.

Similarly , to place a  charge Q_{n} at P_{n} in a field created by (n-1) charges is ,work done to position Q_{n} at P_{n}=V_{n1}Q_{n}+V_{n2}Q_{n}+V_{n3}Q_{n}+.......

\thereforeTotal Work done W_{E} =Q_{2}V_{21}+Q_{3}V_{31}+Q_{3}V_{32}+Q_{4}V_{41}+Q_{4}V_{42}+Q_{4}V_{43}+.... EQN(I)

The total work done is nothing but the Potential energy in the system of charges hence denoted as W_{E},

if charges are placed in reverse order (i.e, first Q_{4} and then Q_{3} and then  Q_{2}  and finally Q_{1} is placed)

work done to place Q_{3} \Rightarrow V_{34}Q_{3}

work done to place Q_{2} \Rightarrow V_{24}Q_{2}+V_{23}Q_{2}

work done to place Q_{1} \Rightarrow V_{14}Q_{1}+V_{13}Q_{1}++V_{12}Q_{1}

Total work done W_{E} =Q_{3}V_{34}+Q_{2}V_{24}+Q_{2}V_{23}+Q_{1}V_{14}+Q_{1}V_{13}+Q_{1}V_{12}+.... EQN(II)

EQN (I)+EQN(II) gives

2W_{E} =Q_{1}(V_{12}+V_{13}+V_{14}+....+V_{1n}) +Q_{2}(V_{21}+V_{23}+V_{24}+....+V_{2n})+Q_{3}(V_{31}+V_{32}+V_{34}+....+V_{3n})+.....

let V_{1}=(V_{12}+V_{13}+V_{14}+....+V_{1n}), V_{2}=(V_{21}+V_{23}+V_{24}+....+V_{2n}) and V_{n}=(V_{n1}+V_{n2}+V_{n3}+....+V_{nn-1}) are the resultant Potentials due to all the charges except that charge.

i.e, V_{1} is the resultant potential due to all the charges except Q_{1}.

2W_{E} = Q_{1}V_{1}+Q_{2}V_{2}+Q_{3}V_{3}+......+Q_{n}V_{n}

W_{E} =\frac{1}{2} \sum_{m=1}^{n}Q_{m}V_{m} Joules.

The above expression represents the Potential Energy stored in the system of n point charges.


W_{E} = \frac{1}{2}\int_{l}\rho _{l}dl. V  Joules

W_{E} = \frac{1}{2}\int_{s}\rho _{s}ds. V Joules

W_{E} = \frac{1}{2}\int_{v}\rho _{v}dv. V Joules  for different types of charge distributions.

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Solved Example problems in Electro Magnetic Theory

  1. Convert Points P(1,3,5)  from Cartesian to Cylindrical and Spherical Co-ordinate system.

Ans. Given P(1,3,5) \fn_cm \Rightarrow x=1,y=3, z=5

Cylindrical :- \fn_cm \phi =tan^{-1}(\frac{y}{x})

                              \fn_cm =tan^{-1}(\frac{3}{1})

                                \fn_cm =75^{o}

Similarly \fn_cm \rho = \sqrt{x^{2}+ y^{2}} \Rightarrow \sqrt{1^{2}+ 3^{2}} = \sqrt{10}=3.16

\fn_cm P(\rho ,\phi ,z)= P(3.16,71.5^{o},5)

Spherical :- \fn_cm r= \sqrt{x^{2}+y^{2}+z^{2}}

                        \fn_cm r= \sqrt{1^{2}+3^{2}+5^{2}}

                      \fn_cm r= \sqrt{35}

                    \fn_cm r=5.91

\fn_cm \theta =tan^{-1}(\frac{\sqrt{x^{2}+y^{2}}}{z})

\fn_cm \theta =tan^{-1}(\frac{\sqrt{1^{2}+3^{2}}}{5})

\fn_cm \theta =32.31^{o}

\fn_cm \phi =tan^{-1}(\frac{y}{x})

\fn_cm \phi =tan^{-1}(\frac{3}{1})

\fn_cm \phi = 75^{o}

\fn_cm P(r,\theta ,\phi ) = P(5.91,32.31^{o},71.5^{o})


Example Problems in Electro Magnetic Theory Wave propagation

  1. A medium like Copper conductor which is characterized by the parameters \bg_black \sigma = 5.8 X 10^{7} Mho's/meter and \epsilon _{r}=1,\mu _{r}=1 uniform plane wave of frequency 50 Hz. Find \alpha ,\beta ,v,\eta  and \lambda.

Ans.  Given \bg_black \bg_black \sigma = 5.8 X 10^{7} Mho's/meter  ,     \bg_black \epsilon _{r}=1,\mu _{r}=1    and \bg_white f= 50 Hz

\bg_white \alpha =? ,\beta =? ,v = ?,\eta =? and \bg_white \lambda =?

Find the Loss tangent \bg_white \frac{\sigma }{\omega \epsilon } = \frac{5.8X 10^{7}}{2 \pi X50X\epsilon _{o}\epsilon _{r}}

                                              \bg_white \bg_white \frac{\sigma }{\omega \epsilon } = \frac{5.8X 10^{7}}{100\pi X\epsilon _{o}}

                                            \bg_white \bg_white \frac{\sigma }{\omega \epsilon } = 2.08 X 10 ^{16}> > 1

So given medium is a Conductor (Copper)

then \bg_white \alpha (or) \beta =\sqrt{\frac{\omega \mu \sigma }{2}}

                         \bg_white =\sqrt{\frac{5.8X10^{7}X2\pi X 50X\mu _{o}}{2}}

                      \bg_white \alpha = 106.99  , \bg_white \beta =106.99.

\bg_white v_{p}=\frac{\omega }{\beta }  \bg_white =\frac{2\pi X50}{106.99}\bg_white =2.936 meters/Sec.

\bg_white \lambda =\frac{2\pi }{\beta }=\frac{2\pi }{106.99}=0.0587 meters.

\bg_white \eta =\sqrt{\frac{j\omega \mu }{(\sigma +j\omega \epsilon )}}

    \bg_white =\sqrt{\frac{jX2\pi X50X\mu _{o}}{(5.8X10^{7}+j2\pi X50X\epsilon _{o})}}

    \bg_white = \sqrt{\frac{j 3.947 X10^{-4}}{(5.8X10^{7}+j 2.78 X10^{-9})}}

    \bg_white = \sqrt{\frac{ 3.947 X10^{-4}\angle 90^{o}}{(5.8X10^{7}\angle -2.74 X 10^{-15})}}

    \bg_white = \sqrt{0.68 X10 ^{-11}}\angle \frac{90-(2.74 X 10^{-5})}{2}

\eta = 2.6 X 10^{-6}\angle 45^{o}.

2. If \bg_white \epsilon _{r}=9,\mu =\mu _{o} for a medium in which a wave with a frequency of \bg_white f= 0.3 GHz is propagating . Determine the propagation constant and intrinsic impedance of the medium when \bg_white \sigma =0.

Ans: Given \bg_white \epsilon _{r}=9,  \bg_white \mu =\mu _{o} , \bg_white f=0.3GHz and \bg_white \sigma =0.

\bg_white \gamma =?,\eta =?

Since \bg_white \sigma =0, the given medium is a lossless Di-electric.

which implies \bg_white \alpha = \frac{\sigma }{2}\sqrt{\frac{\mu }{\epsilon }} =0.

\bg_white \beta = \omega \sqrt{\mu \epsilon }

    \bg_white =2\pi X o.3X10^{9}\sqrt{\mu _{o}X9\epsilon _{o}}

    \bg_white = 18.86.

\bg_white \eta = \sqrt{\frac{\mu }{\epsilon }}

\bg_white \eta = \sqrt{\frac{\mu_{o}\mu _{r} }{\epsilon_{o}\epsilon _{r} }}

\bg_white \eta = \sqrt{\frac{\mu_{o} }{9\epsilon_{o} }}

\bg_white \eta = \frac{120\pi }{3}

\bg_white \eta = 40 Ω.


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Maxwell’s Equations in Point (or Differential form) and Integral form

Maxwell’s Equations for time-varying fields in point and Integral form are:
  1. \overrightarrow{\bigtriangledown }X\overrightarrow{H}=\overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}      \Rightarrow \oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}+\int_{s}\frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}.
  2. \overrightarrow{\bigtriangledown }X\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}       \Rightarrow \oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial \overrightarrow{B}}{\partial t}.\overrightarrow{ds} . 
  3. \overline{\bigtriangledown }.\overrightarrow{D}=\rho _{v}            \Rightarrow \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho _{v}dv.
  4. \overrightarrow{\bigtriangledown }.\overrightarrow{B}=0      \Rightarrow \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0.

The 4 Equations above are known as Maxwell’s Equations. Since Maxwell contributed to their development and establishes them as a self-consistent set.  Each differential Equation has its integral part. One form may be derived from the other with the help of Stoke’s theorem (or) Divergence theorem.

word statements of the field Equations:-

A word statement of the field Equations is readily obtained from their mathematical statement in the integral form.

1.\overrightarrow{\bigtriangledown }X\overrightarrow{H}=\overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t} \Rightarrow \oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}+\int_{s}\frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}.

i.e, The magneto motive force (\because \oint_{l}\overrightarrow{H}.\overrightarrow{dl}\rightarrow is m.m.f)around a closed path is equal to the conduction current plus the time derivative of the electric displacement through any surface bounded by the path.

 2. \overrightarrow{\bigtriangledown }X\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}\Rightarrow \oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial \overrightarrow{B}}{\partial t}.\overrightarrow{ds}.

The electro motive force (\because \oint_{l}\overrightarrow{E}.\overrightarrow{dl}\rightarrow is e.m.f)around a closed path is equal to the time derivative of the magnetic displacement through any surface bounded by the path.

3.\overrightarrow{\bigtriangledown }.\overrightarrow{D}=\rho _{v}  \Rightarrow \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho _{v}dv.

The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume.

4.    \overrightarrow{\bigtriangledown }.\overrightarrow{B}=0  \Rightarrow \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0.

The net magnetic flux emerging through any close surface is zero.

the time-derivative of electric displacement is called displacement current. The term electric current is then to include both conduction current and displacement current. If the time-derivative of electric displacement is called an electric current, similarly \frac{\partial \overrightarrow{B}}{\partial t} is known as magnetic current, e.m.f as electric voltage and m.m.f as magnetic voltage.

the first two Maxwell’s Equations can be stated as 

  1. The magnetic voltage around a closed path is equal to the electric current through the path.
  2. The electric voltage around a closed path is equal to the magnetic current through the path.
Maxwell’s Equations for static fields in point and Integral form are:

Maxwell’s Equations of static-fields in differential form and integral form are:

  1. \overrightarrow{\bigtriangledown } X\overrightarrow{H}=\overrightarrow{J}        \Rightarrow \oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}.
  2. \overline{\bigtriangledown } X\overrightarrow{E}=0           \Rightarrow \oint_{l}\overrightarrow{E}.\overrightarrow{dl}=0.
  3. \overline{\bigtriangledown }.\overrightarrow{D} = \rho _{v}            \Rightarrow \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho _{v}dv.
  4. \overline{\bigtriangledown }.\overrightarrow{B} = 0             \Rightarrow \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0.

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Inconsistensy in Ampere’s law (or) Displacement Current density

Faraday’s experimental law has been used to obtain one of Maxwell’s equations in differential form \overrightarrow{\bigtriangledown }X \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} , which shows that a time-varying Magnetic field produces an Electric field.

From Ampere’s  Circuital law which is applicable to Steady Magnetic fields

\overrightarrow{\bigtriangledown } X \overrightarrow{H}=\overrightarrow{J}

By taking divergence of Ampere’s law the Ampere’s law is not consistent with time-varying fields

\overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown } X \overrightarrow{H})=\overrightarrow{\bigtriangledown }.\overrightarrow{J}

\overrightarrow{\bigtriangledown } . \overrightarrow{J}=0 ,since \overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown } X \overrightarrow{H})=0---------Equation(1)

the divergence of the curl is identically zero which implies \overrightarrow{\bigtriangledown }.\overrightarrow{J}=0------Equation(2), but from the continuity equation \overrightarrow{\bigtriangledown }.\overrightarrow{J} = -\frac{\partial \rho _{v}}{\partial t}-------Equation(3) which is not equal to zero, as \frac{\partial \rho _{v}}{\partial t}\neq 0 is an unrealistic limitation(i.e we can not assume \frac{\partial \rho _{v}}{\partial t} as zero) .

\therefore to make a compromise between the above two situations we must add an unknown term \overrightarrow{G} to Ampere’s Circuital law

i.e, \overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\overrightarrow{G}

then by taking the Divergence of the above equation

\overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown }X\overrightarrow{H}) = \overrightarrow{\bigtriangledown }.\overrightarrow{J}+\overrightarrow{\bigtriangledown }.\overrightarrow{G}------Equation(4)

from Equation(1),Equation(4) becomes     \overrightarrow{\bigtriangledown }.\overrightarrow{J}+\overrightarrow{\bigtriangledown }.\overrightarrow{G}=0

\overrightarrow{\bigtriangledown }.\overrightarrow{G}=-\overrightarrow{\bigtriangledown }.\overrightarrow{J}

thus \overrightarrow{\bigtriangledown }.\overrightarrow{G} = \frac{\partial \rho _{v}}{\partial t}---------Equation(5)

from Maxwell’s first Equation \overrightarrow{\bigtriangledown }.\overrightarrow{D}=\rho _{v} 

then Equation (5) becomes \overrightarrow{\bigtriangledown }.\overrightarrow{J} = \frac{\partial }{\partial t} (\overrightarrow{\bigtriangledown }.\overrightarrow{D})

\overrightarrow{\bigtriangledown }.\overrightarrow{G} =\overrightarrow{\bigtriangledown }. \frac{\partial \overrightarrow{D}}{\partial t}

then   \overrightarrow{G} = \frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

This is the equation obtained which does not disagree with the continuity equation. It is also consistent with all other results. This is a second Maxwell’s Equation is time-varying fields so the term \frac{\partial \overrightarrow{D}}{\partial t} has the dimensions of current density Amperes/Square-meter. Since it results from a time-varying electric flux density (\overrightarrow{D} ) , Maxwell termed it as displacement current density \overrightarrow{J_{D}}.

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\overrightarrow{J_{D}}

\overrightarrow{J_{D}}=\frac{\partial \overrightarrow{D}}{\partial t}

up to this point three current densities are there \overrightarrow{J}=\sigma \overrightarrow{E} , \overrightarrow{J}=\rho _{v} \overrightarrow{v} and \overrightarrow{J_{D}}= \frac{\partial\overrightarrow{D} }{\partial t}.

when the medium is Non-conducting medium \overrightarrow{\bigtriangledown }X \overrightarrow{H}=\frac{\partial\overrightarrow{D} }{\partial t}

the total displacement current crossing any given surface is expressed by the surface integral I_{d} = \oint_{s} \overrightarrow{J_{D}}.\overrightarrow{ds}

I_{d} = \oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

from Ampere’s law \oint_{s}(\overrightarrow{\bigtriangledown }X \overrightarrow{H}).\overrightarrow{ds}=\int_{s} \overrightarrow{J}.\overrightarrow{ds} +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\int_{s} \overrightarrow{J}.\overrightarrow{ds} +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I +I_{d}

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Electric Potential (V)

Electric field intensity \overrightarrow{E} can be calculated by using either Coulomb’s law/Gauss’s law . when the charge distribution is symmetric another way of obtaining \overrightarrow{E} is from the electric scalar potential V

Assume a test charge Q_{t} at A in an Electric field, let points A and B are located at r_{A}and r_{B} units from the origin O,from Coulomb’s law the force acting on a test charge Q_{t} is \overrightarrow{F}= Q_{t}\overrightarrow{E}

The work done in moving a point charge Q_{t} along a differential length \overrightarrow{dl} is dW is given by dW = -\overrightarrow{F}.\overrightarrow{dl}

dW = -Q_{t}\overrightarrow{E}.\overrightarrow{dl}

so the total work done in moving a point charge Q_{t} from A to B is W=-Q_{t}\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}

the direction of work done is always opposite to the direction of displacement.

where A is the initial point and B is the final point. Dividing the work done by the charge Q_{t} gives the potential energy per unit charge denoted by V_{AB},this is also known as potential difference between  the two points A and B.

Thus V_{AB} = \frac{W}{Q_{t}}= -\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}

if we take B as initial point and A as final point , then V_{BA} = \frac{W}{Q_{t}}= -\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}----Equation(1)

To derive the expression for V in terms of charge Q and distance r , we can use the concept of Electric field intensity \overrightarrow{E} produced by a charge Q, which is placed at a distance r

i.e, \overrightarrow{E} = \frac{Q}{4\pi \epsilon _{o}r^{2}}\overrightarrow{a_{r}}

from Equation(1) V_{BA} = -\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}

V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}\overrightarrow{a_{r}}.dr \overrightarrow{a_{r}}  since \overrightarrow{dl}=dr.\overrightarrow{a_{r}}

V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}dr

V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}\left [ \frac{-1}{r} \right ]_{r_{A}}^{r_{B}}

V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}[\frac{1}{r_{B}}-\frac{1}{r_{A}}]


similarly, V_{BA}=V_{A}-V_{B} 

where V_{A} and V_{B} are the scalar potentials at the points A and B respectively. If A is  located at \infty with respect to origin ,with zero potential V_{A} =0 and B is located at a distance r with respect to origin. then the work done in moving a charge from  A (infinity) to B is given by 

V_{AB} = \frac{Q}{4\pi \epsilon _{o}r_{B}}  here r_{B} = r 

\therefore V = \frac{Q}{4\pi \epsilon _{o}r} volts.

hence the potential at any point is the potential difference between that point and a chosen point at which the potential is zero. In other words assuming Zero potential at infinity .

The potential at a distance r from a point charge is the work done per unit charge by an external agent in transferring a test charge from infinity to that point.

i.e, V = -\int_{\infty }^{r}\overrightarrow{E}.\overrightarrow{dl}

So a point charge Q_{1} located at a point P with position vector \overline{r_{1}} then the potential at another point Q with a position vector \overline{r} is 

V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}

As like \overrightarrow{E} superposition principle is applicable to V also that is for n point charges Q_{1},Q_{2},Q_{3},Q_{4},........Q_{n} located at points with position vectors  \overline{r_{1}},\overline{r_{2}},\overline{r_{3}},.......\overline{r_{n}}

then the potential at \overline{r} is 

V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}+\frac{Q_{2}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{2}}\right |}+\frac{Q_{3}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{3}}\right |}+........+\frac{Q_{n}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{n}}\right |}


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Propagation of plane EM wave in conducting medium (or) lossy dielectrics

A lossy dielectric medium is one which an EM wave as it propagates losses power owing to imperfect dielectric,that is a lossy dielectric is an imperfect conductor that is a partially conducting medium (\sigma \neq 0) . 

where as a lossless dielectric is a  (\sigma =0) perfect dielectric,then wave equations for conductors are also holds good here 

i.e, \bigtriangledown ^{2}\overrightarrow{E}=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t} +\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

\frac{\partial }{\partial t}= jw

then \bigtriangledown ^{2}\overrightarrow{E}= j\omega \mu \sigma \overrightarrow{E}+\mu \epsilon(j\omega ) ^{2}\overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E}= (\sigma + j \omega \epsilon )j\omega \mu \overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E}= \gamma ^{2} \overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E}- \gamma ^{2} \overrightarrow{E}=0---------Equation (1)

Equation (1) is called helm holtz equation and \gamma is  called propagation constant.

\gamma ^{2} =j\omega \mu (\sigma +j\omega \epsilon )

\gamma ^{2}= j\omega \mu \sigma -\omega ^{2}\mu \epsilon

Since \gamma is a complex quantity it can be expressed as \gamma = \alpha +j\beta

\alpha– is attenuation constant measured in Nepers/meter.

\beta-is phase constant measured in radians/meter.

(\alpha +j\beta ) ^{2}= j\omega \mu \sigma -\omega ^{2}\mu \epsilon

\alpha^{2} +2j\alpha \beta-\beta ^{2} = j\omega \mu \sigma -\omega ^{2}\mu \epsilon

by equating real and imaginary parts separately \alpha ^{2}-\beta ^{2}= -\omega ^{2}\mu \epsilon------Equation(2)

 and 2\alpha \beta =\omega \mu \sigma

\alpha =\frac{\omega \mu \sigma}{2\beta }

 by substituting  \alpha value in the equation (2)   \frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4\beta ^{2}}-\beta ^{2}=-\omega ^{2}\mu \epsilon

{\omega ^{2}\mu ^{2}\sigma ^{2}}-4\beta ^{4}=-4\omega ^{2}\beta ^{2}\mu \epsilon

4\beta ^{4}-4\omega ^{2}\beta ^{2}\mu \epsilon -{\omega ^{2}\mu ^{2}\sigma ^{2}}=0

let \beta ^{2}=t 

4t^{2}-4\omega ^{2}t\mu \epsilon -{\omega ^{2}\mu ^{2}\sigma ^{2}}=0

t^{2}-\omega ^{2}t\mu \epsilon -\frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4}=0

the roots of the above quadratic expression are

t=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}-4(-\frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4})}}{2}

t=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}

\beta ^{2}=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}

\beta =\sqrt{\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}}

\beta =\sqrt{\frac{\omega ^{2}\mu \epsilon (1+ \sqrt{(1+\frac{\sigma }{\omega \epsilon })^{2}})}{2}}




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Electro Magnetic Wave Equation

Assume a Uniform, Homogeneous,linear,isotropic and Stationary medium with Non-zero current i.e, \overrightarrow J_{c}(\sigma \overrightarrow{E}) \neq 0.

When an EM wave is travelling in a conducting medium in which  \overrightarrow J\neq 0. The wave is rapidly attenuated in a conducting medium and in a good conductors, the attenuation is so high at Radio frequencies. The wave penetrates the conductor only to a small depth.

choose the equation \overrightarrow{\bigtriangledown }X \overrightarrow{H} = \overrightarrow{J_{C}}+\overrightarrow{J_{D}} the time-domain representation of it is    \overrightarrow{\bigtriangledown }X \overrightarrow{H} = \overrightarrow{J_{C}}+\frac{\partial \overrightarrow{D}}{\partial t}   

since  \frac{\partial }{\partial t} =j\omega in phasor-notation \overrightarrow{\bigtriangledown }X \overrightarrow{H} = \sigma \overrightarrow{E}+j\omega \epsilon \overrightarrow{E} , \because \overrightarrow{J_{C}}=\sigma \overrightarrow{E} and \overrightarrow{D}=\epsilon \overrightarrow{E} . 

\overrightarrow{\bigtriangledown }X \overrightarrow{H} = \sigma \overrightarrow{E}+\frac{\epsilon\partial \overrightarrow{E}}{\partial t}--------------Equation (1)

 By differentiating the above equation with respect to time\frac{\partial(\overrightarrow{\bigtriangledown } X \overrightarrow{H})}{\partial t} =\sigma \frac{\partial \overrightarrow{E}}{\partial t} + \epsilon \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}--------------Equation (2)

From the Maxwell Equation \overrightarrow{\bigtriangledown } X \overrightarrow{E} =-\frac{\partial \overrightarrow{B}}{\partial t}

By taking curl on both sides \overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E} =- \overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{B}}{\partial t}

\overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E} =- \overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{(\mu H)}}{\partial t}  since \overrightarrow{B} = \mu \overrightarrow{H}

\overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E} =-\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t}) 

By using the vector identity 

\bigtriangledown ^{2}\overrightarrow{E} = \overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }. \overrightarrow{E}) -\overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E} = \overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }. \overrightarrow{E}) -(-\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t}))

\bigtriangledown ^{2}\overrightarrow{E} = \overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }. \overrightarrow{E}) +\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t})

From Equation (2) 

\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t})=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

\bigtriangledown ^{2}\overrightarrow{E}-\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{E})=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

\bigtriangledown ^{2}\overrightarrow{E}=\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{E})+\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

from Maxwell’s equation \overrightarrow{\bigtriangledown }.\overrightarrow{D} = \rho _{v}

\overrightarrow{\bigtriangledown }.\overrightarrow{E} = \frac{\rho _{v}}{\epsilon }

\bigtriangledown ^{2}\overrightarrow{E}=\overrightarrow{\bigtriangledown }(\frac{\rho _{v}}{\epsilon })+\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}---------EquationI

This is called Wave equation (Electric field)for a general medium when {\rho _{v}}\neq 0.

wave equation for Magnetic field of a general  medium is 

\bigtriangledown ^{2}\overrightarrow{H}=\overrightarrow{\bigtriangledown }(\frac{\rho _{v}}{\epsilon })+\mu \sigma \frac{\partial \overrightarrow{H}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{H}}{\partial t^{2}}---------EquationII.

Case 1:-  wave equation for a conducting medium

for a conductor the net charge inside an isolated conductor is \rho _{v}=0, then the wave equation for a conducting medium (\sigma \neq 0) is

\bigtriangledown ^{2}\overrightarrow{E}=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}---------EquationI

\bigtriangledown ^{2}\overrightarrow{H}=\mu \sigma \frac{\partial \overrightarrow{H}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{H}}{\partial t^{2}}---------EquationII

The above equations are known as wave equations for conducting medium and are involving first and second order time derivatives, which are well known equations for damped(or) attenuated waves in absorbing medium of homogeneous, isotropic such as metallic conductor.

Case 2:-  Wave equation for free space/ Non-conducting medium/loss-less medium/Perfect Di-electric medium

The conditions of free space are \rho =0,\sigma =0,\overline{J} =0,\mu =\mu _{o} and \epsilon =\epsilon _{o}

By substituting the above equations in general wave equations, the resulting wave equations for non-conducting medium are

\bigtriangledown ^{2}\overrightarrow{E}=\mu_{o} \epsilon_{o} \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}---------EquationI

\bigtriangledown ^{2}\overrightarrow{H}=\mu_{o} \epsilon_{o} \frac{\partial ^{2}\overrightarrow{H}}{\partial t^{2}}---------EquationII.

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Electric field due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density \rho _{L} C/m and lies on Z-axis from -\infty to +\infty.

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are (0,\rho ,0) ( a point on y-axis) and assume dQ is a small differential charge confirmed to a point  M (0,0,Z) as co-ordinates.

\therefore dQ produces a differential field \overrightarrow{dE} 

\overrightarrow{dE}=\frac{dQ}{4\pi \epsilon _{o}R^{2}}\widehat{a_{r}}

the position vector \overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }} and the corresponding unit vector \widehat{a_{r}} =\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{\rho ^{2}+Z^{2}}}

\therefore \overrightarrow{dE} =\frac{dQ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

therefore \overrightarrow{dE} =\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

then the Electric field strength \overrightarrow{E} produced by the infinite line charge distribution \rho _{L} is 

\overrightarrow{E} = \int \overrightarrow{dE}

\overrightarrow{E} = \int_{z=-\infty }^{\infty }\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

to solve this integral  let Z= \rho \tan \theta \Rightarrow dZ=\rho \sec ^{2}\theta d\theta

as Z\rightarrow -\infty \Rightarrow \theta \rightarrow \frac{-\pi }{2}

Z\rightarrow \infty \Rightarrow \theta \rightarrow \frac{\pi }{2}

\therefore \overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{(\rho ^{2}+\rho ^{2}\tan ^{2}\theta )^{\frac{3}{2}}})  

\overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{\rho ^{3}\sec ^{3}\theta })  

\overrightarrow{E} = \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }(\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} -\sin \theta \overrightarrow{a_{z} }d\theta +\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \cos\theta d\theta \overrightarrow{a_{\rho }})

\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }(0+2\overrightarrow{a_{\rho }})

\therefore \overrightarrow{E}= \frac{\rho _{L}}{2\pi\epsilon _{o}\rho }\overrightarrow{a_{\rho }} Newtons/Coulomb.

\overrightarrow{E} is a function of \rho only, there is no \overrightarrow{a_{z}} component and \rho is the perpendicular distance from the point P to line charge distribution \rho _{L}.



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Gauss Divergence Theorem (or) Application of Gauss’s law to a differential volume element

Gauss Divergence theorem:-

A vector field changes from point to point throughout space, this change can be identified by two things Divergence and curl,the second is curl, which will be examined when magnetic fields are discussed, so Divergence , which is a scalar function and is somewhat similar to derivative of a function will be discussed in Gauss-Divergence theorem.

Generally, when the divergence ≠ 0 ⇒ that there exists sources or sinks in a particular region.

If divergence is positive ⇒ a source                                                                                    If divergence is negative ⇒ a sink

In static electric fields, there is a correspondence between positive divergence, sources and positive electric charge ‘Q’, since electric flux ψ by definition originates on a positive charge, thus a region which consists of positive charges contains the source of flux ψ.

∴ The divergence of electric flux density \overrightarrow{D} will be positive in that region.Divergence of a vector field \overrightarrow{A}/\overrightarrow{D} at a point ‘P’ is defined by 

div\overrightarrow{A}= \lim_{\Delta v->0}\frac{\oint \overrightarrow{A}.\overrightarrow{ds}}{\Delta v}            or      div\overrightarrow{D}= \lim_{\Delta v->0}\frac{\oint \overrightarrow{D}.\overrightarrow{ds}}{\Delta v}

here, the integration is over the surface of an infinitesimal volume \Delta v that shrinks at point P

Gauss-Divergence theorem:- (statement)

for a continuously differentiable vector field the net outward flux from a closed surface equals the volume integral of the divergence throughout the region bounded by that surface.

i.e, \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}(\overrightarrow{\bigtriangledown}. \overrightarrow{D})dv   where \overrightarrow{D} be the vector field.

It relates volume integral with surface integral.

Proof of Divergence theorem:-

Now consider a slightly different type of situation to which Gauss’s law is to be applied, that is on the surface \overrightarrow{D} is not constant (or) zero and this surface doesn’t have any symmetry at all. For such type of surface the change in \overrightarrow{D} can be represented by Taylor series expansion by assuming \overrightarrow{D} as constant.

i.e, f(x+\frac{\Delta x}{2})=f(x)+\frac{\frac{\Delta x}{2}}{1!}{f}'(x)+\frac{\frac{\Delta x}{2}^{2}}{2!}{f}''(x)+..... over the surface.

This change in \overrightarrow{D} can be expressed by only first two terms 

i.e, f(x+\frac{\Delta x}{2})=f(x)+\frac{\frac{\Delta x}{2}}{1!}{f}'(x)   

for the differential volume element \large \Delta x =dx,\Delta y=dy and \Delta z=dz

Let us consider a point ‘P’ which is located at the center of the volume element, which is a differential volume with an assumption that \overrightarrow{D} is almost constant over this small volume dv and at P \overrightarrow{D} is given as 

\overrightarrow{D}=D_{xo}\overrightarrow{a}_{x} +D_{yo}\overrightarrow{a}_{y}+D_{zo}\overrightarrow{a}_{z}

Now, from Gauss’s law \oint_{s}\overrightarrow{D}.\overrightarrow{ds} =Q_{enclosed} , the surface integral is divided into ‘6’ integrals as \oint_{s}\overrightarrow{D}.\overrightarrow{ds} =\int_{front}+\int_{back}+\int_{left}+\int_{right}+\int_{top}+\int_{bottom}

\int_{front} = \overrightarrow{D}_{front}.\overrightarrow{d s}_{front}

\large \overrightarrow{D}_{front} = \overrightarrow{D}_{xfront}\overrightarrow{a}_{x} +\overrightarrow{D}_{yfront}\overrightarrow{a}_{y} +\overrightarrow{D}_{zfront}\overrightarrow{a}_{z}

\large \overrightarrow{d s}_{front} = \Delta y\Delta z \overrightarrow{a}_{x}

\large \overrightarrow{D}_{front}.\overrightarrow{ds}_{front}= (D_{xfront}\overrightarrow{a}_{x}+D_{yfront}\overrightarrow{a}_{y}+D_{zfront}\overrightarrow{a}_{z}).dydz\overrightarrow{a}_{x}

                                   \large = D_{xfront}dydz

now the component \large D_{xfront}=D_{xo}+\frac{dx}{2}\frac{\partial D_{x}}{\partial x}

\large \therefore \int_{front}=(D_{xo}+\frac{dx}{2}\frac{\partial D_{x}}{\partial x})dydz

simillarly,  \large \therefore \int_{back}=(D_{xo}-\frac{dx}{2}\frac{\partial D_{x}}{\partial x})(-dydz)

\large \int_{front}+\int_{back} = dxdydz \frac{\partial D _{x}}{\partial x}

then the remaining two integral values \large \int_{left}+\int_{right} = dxdydz \frac{\partial D _{y}}{\partial y}

\large \int_{top}+\int_{bottom} = dxdydz \frac{\partial D _{z}}{\partial z}

\large \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=dxdydz \frac{\partial D _{x}}{\partial x}+dxdydz \frac{\partial D _{y}}{\partial y}+ dxdydz \frac{\partial D _{z}}{\partial z} —————>Equation(1)

According to the definition of div \large \overrightarrow{D}

\large \lim_{\Delta v->0}\frac{\oint \overrightarrow{D}.\overrightarrow{ds}}{\Delta v}=div\overrightarrow{D}=\overrightarrow{\bigtriangledown }.\overrightarrow{D}

from Equation(1) \large \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=dv( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

by dividing the above equation with dv=Δv

\large \frac{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}{dv}=( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

apply the limit on both sides

\large \lim_{dv->0}\frac{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}{dv}=\lim_{dv->0}( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

\large \overrightarrow{\bigtriangledown }.\overrightarrow{D}= div \overrightarrow{D}=( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})

hence proved.

The divergence of the  flux density \large \overrightarrow{D} (vector ) is the flow of flux from a small closed surface per unit volume as the volume shrinks to zero \large (\lim_{dv->0}).


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Electricfield Intensity-E

Electric field Intensity(\overrightarrow{E}):-

Let us suppose a point charge Q is placed somewhere in space and if any other charge q is brought near to it, q experiences a force on Q and vice-versa. Thus there exists a region around a charge in which it experience a force on any other charge located in that region.

∴ The region around a charge distribution is called as Electric field produced by that charge distribution.

The electric field intensity (or) Electric field strength is defined as the force  per unit charge (test charge)

i.e, \overrightarrow{E} = \frac{\overrightarrow{F}}{q} Newtons/Coulomb —-> Equation (1)

∴ The expression for    \overrightarrow{E_{at P}}=\frac{Q}{4\pi \epsilon _{o}R^{2}}\widehat{a_{R}} 

q is small test charge +Q is a positive charge placed in free space. \overrightarrow{E} is the electric field produced around +Q charge, then after placing a small test charge q in a field \overrightarrow{E} ,\overrightarrow{E} exerts some force on this test charge q , given by 

F = E q

and simillarly +Q and q experiences force on each other given by F = \frac{Qq}{4\pi \epsilon r^{2}}—>Equation(2) (magnitude)

Force (as a vector) \overrightarrow{F} = \frac{Qq}{4\pi \epsilon r^{2}}\widehat{a_{R}} 

from equations (1) and (2)   F   =  F  (magnitudes)

                                                 Eq = \frac{Qq}{4\pi \epsilon r^{2}}

                                                  E= \frac{Q}{4\pi \epsilon r^{2}}—> Scalar magnitude of \overrightarrow{E} produced by the charge Q.

where as \overrightarrow{E}= \frac{Q}{4\pi \epsilon r^{2}}\widehat{a_{R}} which acts along the direction of Coulomb’s force \overrightarrow{F}.

Types of charge distributions:-

In order to find out the electric field strength due to different types of charge distributions, first of all one must know how many types of charge distributions are there? the positive and negative charges can be distributed into 3 types of distributions.


  1. charge is conserved , i.e, charge is neither be created nor destroyed.
  2. charges are surrounded by electric and magnetic fields.

Point charge distribution:-

The name itself indicates that the charge confined to a point is known as point charge distributions. In  practical point charges my not exists and point charge does not occupy any space.

Example:- an electron with a charge of 1.6X10-19 C is a point charge.

Line charge distribution(ρL):-

If the charge is distributed along the length of the line, then it is known as line charge distribution. It may be a uniform (or) non-uniform distribution as shown in the figures.

If the charges are distributed uniformly along the line then it is a uniform charge distribution ρL is constant through out the line, otherwise it is Non-uniform type.

The line charge density \rho _{L}=\lim_{\Delta l->0}\frac{\Delta Q}{\Delta l}

\rho _{L}=\frac{dQ}{dl}=\frac{Q}{l} Coulomb/m—>   Q=\int_{l}\rho _{l}dl Coulomb

\rho _{L}  is defined as the charge per unit-length.

Surface charge distribution(ρs):-

If the charge is distributed uniformly over a two dimensional surface. Then it is called uniform surface charge distribution otherwise non-uniform.

then the surface charge density \rho _{s}= \lim_{\Delta s->0}\frac{\Delta Q}{\Delta s}

\rho _{s}=\frac{dQ}{ds}=\frac{Q}{S} c/m^{2}

Q=\int_{s}\rho _{s}ds Coulomb

ρs is defined as charge per unit surface area and is measured in terms of C/m2.

Volume charge distribution (ρv):-

If the charge is distributed uniformly in a volume then it is called as uniform volume distribution. Sphere represents a volume here

\rho _{v}=\lim_{\Delta v->0}\frac{\Delta Q}{\Delta v}

\rho _{v}=\frac{dQ}{dv}=\frac{Q}{V} c/m^{3}

Q=\int_{v}\rho _{v}dv Coulombs


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Gauss’s law and its applications

Gauss’s law:-

Statement:- Gauss’s law states that the total flux coming out of a closed surface is equal to the net charge enclosed by that surface.

i.e, \oint_{s}\overrightarrow{D}.\overrightarrow{ds}= Q_{enclosed}      —–>               i.e, \psi _{e} = Q_{enclosed}


Consider a closed surface of any shape with charge distribution as shown in the given figure. Now this surface is enclosed by a charge of Q coulombs , then a flux of Q coulombs will pass through the enclosing surface.

Now divide the entire area S into small pieces of differential areas ds or Δs.

Now at point P , the flux density is Ds and is direction is as shown in the figure. The surface S is chosen is an irregular surface and \overrightarrow{D} is also not uniform. i.e, \overrightarrow{D} direction as well as it’s magnitude is going to change from point to point. 

now for the surface area ds the normal vector to the surface is \overrightarrow{ds_{n}}= ds.\overrightarrow{a_{n}}the direction of \overrightarrow{D} at a point P on the surface ds is making an angle θ \overrightarrow{ds_{n}}, then flux density at point ‘P’ is D = \frac{d\psi }{ds}

d\psi = D.ds—>equation 1 

to get maximum flux out of the surface \overrightarrow{D} and \overrightarrow{ds_{n}} should be in the same direction, there is a need to find out the component of \overrightarrow{D} along \overrightarrow{ds_{n}} is D_{s}cos\theta = D_{s normal}

then equation 1 becomes d\psi = D_{snormal} ds

d\psi =D _{s}cos\theta ds —> d\psi = \overrightarrow{D}.\overrightarrow{ds}

Total flux is \psi = \oint_{s}d\psi =\oint_{s}\overrightarrow{D}.\overrightarrow{ds}

\therefore \psi =\oint_{s}\overrightarrow{D}.\overrightarrow{ds}

∴ Total flux \psi = net charge enclosed Q

\therefore Q _{enclosed}=\oint_{s}\overrightarrow{D}.\overrightarrow{ds}

If there are n number of charges Q1, Q2,Q3 …..Qn then Q= \sum Q_{n}

i. for a line charge distribution Q=\int_{l}\rho _{l}dl                                                                      ii. for a surface charge distribution Q=\int_{s}\rho _{s}ds                                                            iii. for a volume charge distribution Q=\int_{v}\rho _{v}dv.

Closed Gaussian surface:-

The Gauss’s  law is used to find out  \overrightarrow{E} or \overrightarrow{D} for symmetrical charge distributions and is used to find out \psi or Q of any closed surface.

  1. \overrightarrow{D} is every where either normal (or) tangential to the closed surface, that is θ = π/2 or θ = 0o / π so that \overrightarrow{D}.\overrightarrow{ds} is maximum or zero.
  2. \overrightarrow{D} is constant over the portion of the closed surface for which \overrightarrow{D}.\overrightarrow{ds} is not zero.

To apply Gauss’s law to a charge distribution, surface is required to be a Gaussian surface that encloses the  charge distribution or charged body.


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EMT-Electrostatics-coulomb’s law


Electrostatics deals with static electric fields ( charges which are at rest). 

i.e, Electric fields which are independent of time, these are the fields produced by the charges at rest. A charge can be either concentrated at a point or distributed in some fashion like line, surface, volume and the charge distribution is assumed to be constant with respect to time.

Coulomb’s law:-

Let us suppose there exists two charged bodies placed apart at a distance ‘R’ as shown in the figure

then there exists a force between the two charges, if the two charges are like charges , the force is repulsive in nature. where as  for unlike charges the force is attractive in nature, that is the force is either the force of attraction (or) the force of repulsion which is given by 

F \propto Q_{1}Q_{2}

      \propto \frac{1}{R^{2}}


The force of attraction or repulsion between the two charged bodies                      i. is directly proportional to the product of the two charges.                                   ii. is inversely proportional to the square of the distance between them.           iii. and acts along the line joining the two point charges.

i.e, F =\frac{k Q_{1}Q_{2}}{R^{2}}    where k  is constant of proportionality

    F = \frac{Q_{1}Q_{2}}{4\pi \epsilon R^{2} }  , k=\frac{1}{4\pi \epsilon }

where \epsilon is the absolute permittivity of the medium given by \epsilon = \epsilon _{o}\epsilon _{r}

here \epsilon _{o} is free space permittivity

\epsilon _{r} is the relative permittivity of the medium.

\epsilon _{o} =8.854 X 10^{-12}F/m     or    \epsilon _{o} = \frac{10^{-9}}{36\pi } F/m

k= \frac{1}{4\pi \epsilon _{o}}= 9X10^{9}m/F

permittivity (or) capacitivity (∈) :-

This is defined as the ability (or) Capacity to store electrical energy                     r : 8 to 9 for alumina.                                                                                                             r : 2 to 3 for Teflon fibre glass. 

the force is scalar one in the previous equation, Now the vector form is 

\overrightarrow{F} = \frac{Q_{1}Q_{2}}{4\pi \epsilon _{o}R^{2}}\hat{a_{R}}   where \hat{a_{R}} is the unit vector in the direction of R

If there exists two charges, Q1 and Q2 then force acting on 1 due to 2 is given by  {\overrightarrow{F}_{12}} = \frac{Q_{1}Q_{2}}{4\pi \epsilon _{o}R_{21}^{2}}\hat{a_{21}}

simillarly Force acting on 2 by 1 is given by 

\overrightarrow{F}_{21}= \frac{Q_{1}Q_{2}}{4\pi \epsilon _{o}R_{12}^{2}}\hat{a_{12}}

\therefore \overrightarrow{F_{12}}=-\overrightarrow{F_{21}} 

both are equal in magnitude, they differ in their directions.

Q1. Two point charges 0.7mC and 4.9uC are situated in free space at (2,3,6) and(0,0,0) . Calculate the force acting on the 0.7mC charge.

Ans:  Let Q1 = 0.7mC     and Q2 = 4.9uC

Force acting on 1 by 2 is

\overrightarrow{F}_{12} = \frac{Q_{1}Q_{2}}{4\pi\epsilon _{o}R_{21}^{2}}\hat{a_{21}} = \frac{0.7X10^{-3}4.9X10^{-6}(2\overrightarrow{a_{x}}+3\overrightarrow{a_{y}}+6\overrightarrow{a_{z}})}{4\pi \epsilon _{o}X7^{2}(\sqrt{4+9+36})}

\overrightarrow{F}_{12}= 0.18\overrightarrow{a_{x}}+0.27\overrightarrow{a_{y}}+0.54\overrightarrow{a_{z}} Newtons

Force due to number of charges:-

Imagine a situation when there exists more than two charges, then each will experience a force on the other, then the resultant force on any charge can be obtained by the principle of superposition (i.e linear addition).

the total force on Qo in such case is vector sum of all forces acting on Qo by each of the charges  Q1 , Q2  & Q3

force acting on Qo due to Q1 is

\overrightarrow{F}_{o1}= \frac{Q_{o}Q_{1}}{4\pi \epsilon _{o}R_{1o}^{2}}\hat{a_{1o}} 

force acting on Qo due to Q2 is

\overrightarrow{F}_{o2}= \frac{Q_{o}Q_{2}}{4\pi \epsilon _{o}R_{2o}^{2}}\hat{a_{2o}}  

force acting on Qo due to Q3 is

\overrightarrow{F}_{o3}= \frac{Q_{o}Q_{3}}{4\pi \epsilon _{o}R_{3o}^{2}}\hat{a_{3o}}

then the Resultant force is 

\overrightarrow{F}=\overrightarrow{F_{o1}}+\overrightarrow{F_{o2}}+\overrightarrow{F_{o3}} , This is for taking four charges into account, if there exists ‘n’ number of charges, then the force action on Qo by the remaining (n-1) charges on it is given by

then \overrightarrow{F}=\frac{Q_{o}}{4\pi\epsilon _{o}}\sum_{i=1}^{n}\frac{Q_{i}\hat{a_{io}}}{R_{io}^{2}}.


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Hello world! this blog is for B.TECH e.c.e students. This is used to provide information about various subjects of E.C.E

Hello world. This blog is especially for students who are pursuing Engineering degree under stream of Electronics and Communications. The main focus of this is to provide useful information as well as basics related to certain subjects which are very useful to clear competitive exams.Please Share your  valuable opinion on this blog in the comment section.ece