Question

Two free charges +4uc and - 2uc are separated by a distance of 1m in air at what point on the line joining the electrode potential become zero

Answer 1

Given:

2 free charges of magnitude +4 microcoulomb and -2 microcoulomb are separated by a distance of 1 metre in air.

To find:

Point on the line joining the charges where the Electrostatic Potential will be zero.

Calculation:

Let the position of zero potential be x cm from 4 microcoulomb charge , then it will be (100-x) cm from the -2 microcoulomb charge.

The electrostatic potential created by both the charges will be equal and opposite at a particular point. Hence the net Potential will become zero.

$\therefore \dfrac{1}{4\pi\epsilon_{0}} \bigg(\dfrac{4 \mu }{x} \bigg) = \dfrac{1}{4\pi\epsilon_{0}} \bigg(\dfrac{2 \mu }{100 - x} \bigg)$

Cancelling the common terms:

$= > \cancel{ \dfrac{1}{4\pi\epsilon_{0}} } \bigg(\dfrac{4 \cancel\mu }{x} \bigg) = \cancel{\dfrac{1}{4\pi\epsilon_{0}}} \bigg(\dfrac{2 \cancel\mu }{100 - x} \bigg)$

$= > \dfrac{4}{x} = \dfrac{2}{100 - x}$

$= > 400 - 4x = 2x$

$= > 6x = 400$

$= > x = \dfrac{400}{6}$

$= > x = 66.67 \: cm$

So the zero potential point is located 66.67 cm from +4 microcoulomb charge and 33.33 cm from the -2 microcoulomb charge.