Question

infinite number of particles each of charge q are placed on x-axis.
They are at x= 1,2,4,8....etc. What is the force experienced by another particle of q° which is placed at the origin.

Pls ANS with all the _steps_

FIRST TO ANSWER FULLY WILL BE THE BRAINLIEST ​

Answer 1

I think this will help you

Answer 2

Answer:

Given:

q° is placed at origin. And infinite number of q charges are placed at x = 1,2,4,8,.....

To find:

Net force experienced by q°

Calculation:

The net force will be the vector sum of the forces due to the all the charges placed at different positions on the X axis.

$f = \dfrac{kqq \degree}{ {1}^{2} } + \dfrac{kqq \degree}{ {2}^{2} } + \dfrac{kqq \degree}{ {4}^{2} } + ... \infty$

Taking kqq° common :

$= > f = kqq \degree \bigg \{ \frac{1}{ {1}^{2} } + \frac{1}{ {2}^{2} } + \frac{1}{ {4}^{2} } ... \infty \bigg \}$

$= > f = kqq \degree \bigg \{ \frac{1}{ 1 } + \frac{1}{4 } + \frac{1}{ 16 } ... \infty \bigg \}$

So we can understand this as a Geometric Progression with First term as 1 (a) and common ratio as ¼ (r).

Applying formula for infinite sum in GP

$= > f = kqq \degree \bigg \{ \red{ \dfrac{a}{1 - r}} \bigg \}$

$= > f = kqq \degree \bigg \{ \dfrac{1}{1 -( \frac{1}{4} )} \bigg \}$

$= > f = kqq \degree \bigg \{ \dfrac{1}{( \frac{3}{4} )} \bigg \}$

$= > f = kqq \degree \times \dfrac{4}{3}$

$= > f = \dfrac{4kqq \degree}{3}$

So final answer :

$\boxed{ \green{ \sf{ \huge{ \bold{f = \dfrac{4kqq \degree}{3}}}}}}$