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}}.

 

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Author: Lakshmi Prasanna

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.