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

force acting on Q_o 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 Q_o 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 Q_o 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 Q_o by the remaining (n-1) charges on it is given by

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

Note: There is a rating embedded within this post, please visit this post to rate it.

Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems. View all posts by Lakshmi Prasanna Ponnala

Electrostatics-coulomb’s law

Electrostatics:-

Like this:

Related

Author: Lakshmi Prasanna Ponnala

Leave a ReplyCancel reply

Electrostatics:-

Like this:

Related

Author: Lakshmi Prasanna Ponnala

Leave a ReplyCancel reply

Discover more from ece4uplp