GATE problems in EMT

pb. An Electro Magnetic Wave Propagates through a loss less insulator with a velocity 1.5 X 10^{10} Cm/sec . Calculate the electric magnetic properties of the insulator if its intrinsic impedance is 90\pi Ohms.

Ans:- Given loss less insulator  \Rightarrow \alpha =0

v_{p} =1.5 X 10^{10} Cm/Sec

v_{p} = \frac{1}{\sqrt{\mu _{o}\epsilon _{o}\mu _{r}\epsilon _{r}}}

1.5 X 10^{10}= \frac{1}{\sqrt{\mu _{o}\epsilon _{o}\mu _{r}\epsilon _{r}}}

1.5 X 10^{10}= \frac{3 X 10^{10}}{\sqrt{\mu _{r}\epsilon _{r}}}

{\sqrt{\mu _{r}\epsilon _{r}}} = 2 ------------EQNI

from \eta =\sqrt{\frac{\mu _{o}\mu _{r }}{\epsilon _{o}\epsilon _{r }}}

90\pi = \sqrt{\frac{\mu _{o}\mu _{r }}{\epsilon _{o}\epsilon _{r }}}

90\pi = \sqrt{\frac{\mu _{r }}{\epsilon _{r }}} . 120 \pi

\sqrt{\frac{\mu _{r }}{\epsilon _{r }}} =\frac{3}{4}--------EQNII

from Equations I and II \mu _{r} = \sqrt{\frac{\mu _{r }}{\epsilon _{r }}} \sqrt{\mu _{r }\epsilon _{r}}

\mu _{r} = 2 X \frac{3}{4}

\mu _{r} = 1.5

\sqrt{\mu _{r }\epsilon _{r }} = 2

\sqrt{1.5 \epsilon _{r }} = 2

\epsilon _{r} = 2.66

[ratings]

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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.

 

[ratings]

 

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.

[ratings ]

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}

[ratings ]

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}}]

V_{AB}=V_{B}-V_{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 |}

 

[ratings ]

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}}

similarly,

 

 

[ratings ]

 

 

 

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.

[ratings ]

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

 

 

[ratings ]

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}).

 

[ratings ]

 

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.

Properties:-

  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

 

[ratings ]

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}

Proof:- 

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 w.r.to \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.

 

[ratings ]

EMT-Electrostatics-coulomb’s law

Electrostatics:-

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}}

Statement:-

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}}.

 

[ratings ]

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