Question

Two charges of -5×10^-9 C and -1.5×10^-9 C are separated by a distance of 43cm. Find the equilibrium position for a third charge of +1.1×10^-8 C by identifying its distance from the first charge q1.

Answer 1

Solution

In this question we have given

Charge,$q_{1}=-5\times 10^-9C$

Charge,$q_{2}=-1.5\times 10^-9C$

Charge,$q_{3} =1.1\times 10^-8C$

Distance between Charge $q_{1}$ and $q_{2}$,d=43cm=.43m

Let the distance between charge $q_{1}$ and $q_{3}$, is x

Therefore,distance between charge $q_{2}$ and $q_{3}$, will be=(.43-x)

We know By Couloms law, Force on Charge $q_{3}$ due to charge $q_{1}$ which are seperated by distance d is given by following equation

$F_{1}=\frac{k\times q_{1}q_{2}}{d^2}$............(1)

Here, K=$9\times 10^9 Nm^2C^-2$

Therefore,

Force on Charge $q_{3}$ due to charge $q_{1}$

$F_{1}=-\frac{k\times 5\times 10^-9C\times 1.1\times 10^-8C}{x^2}$..(2)

Similarly,

Force on Charge $q_{3}$ due to charge $q_{2}$

$F_{2}=-\frac{k\times 1.5\times 10^-9C\times 1.1\times 10^-8C}{(.43-x)^2}$..(2)

In equilibrium condition,

$F_{1} =F_{2}$

$-\frac{k\times 5\times 10^-9C\times 1.1\times 10^-8C}{x^2}=-\frac{k\times 1.5\times 10^-9C\times 1.1\times 10^-8C}{(.43-x)^2}$

$\frac{5\times 10^-9C}{x^2}=\frac{1.5\times 10^-9C}{(.43-x)^2}$

$5\times 10^-9C \times (.43-x)^2=1.5\times 10^-9C\times (x)^2$

$3.5x^2-4.3x+.9245=0$

on solving we get

x=.277m

0r x=28cm

The distance between charge $q_{1}$ and $q_{3}$, is 28cm