Question

find the position of neutral point along the line joining the two point charge Qa=-40mC and Qb=20mC which are 2cm apart in air ​

Answer 1

Given:

A line joins the two point charge Qa=-40mC and Qb=20mC which are 2cm apart in air .

To find:

Position of neutral point.

Calculation:

Neutral point is that position on the line where the net electrostatic field intensity is zero.

Since the charges are opposite in nature, the neutral point to be located near to the smaller charge.

So , let 2 mC be x cm from neutral point , then 4 mC will be located (x + 2) cm from neutral point.

$\therefore \: \dfrac{k \times 20 \times {10}^{ - 3} }{ {x}^{2} } = \dfrac{k \times 40 \times {10}^{ - 3} }{ {(2 + x)}^{2} }$

$= > \dfrac{1}{ {x}^{2} } = \dfrac{2}{ {(2 + x)}^{2} }$

$= > \dfrac{2 + x}{ x } = \sqrt{2}$

$= > 2 + x = \sqrt{2} x$

$= > ( \sqrt{2} - 1)x = 2$

$= > x = \dfrac{2}{ \sqrt{2} - 1 } \: cm$

So , neutral point is located

2/(√2-1) cm from 2mC charge.