Question

Two fixed charges 9q and q at a distance 6m apart at what pointbetween the charges the net electrostatic force will be.0

Answer 1

Given that,

First charge = q

Second charge = 9q

Distance = 16 m

Let the third charge q be placed at a distance x from q and hence its distance from 9q and (16−x) cm.

We need to calculate the electric force for first charge

Using formula of electrostatic force

$F_{1}=\dfrac{kq_{1}Q}{r^2}$

Put the value into the formula

$F_{1}= \dfrac{kqQ}{x^2}$

We need to calculate the electric force for second charge

Using formula of electrostatic force

$F_{2}=\dfrac{kq_{2}Q}{r^2}$

Put the value into the formula

$F_{2}=\dfrac{k9qQ}{(16-x)^2}$

We need to calculate the value of x

Using formula of net electrostatic force

$F_{1}-F_{2}=0$

$F_{1}=F_{2}$

Put the value into the formula

$\dfrac{kqQ}{x^2}=\dfrac{k9qQ}{(16-x)^2}$

$\dfrac{1}{x^2}=\dfrac{3}{(16-x)^2}$

$\dfrac{1}{x}=\dfrac{3}{16-x}$

$16-x=3x$

$4x=16$

$x=\dfrac{16}{4}$

$x=4\ cm$

We need to calculate the distance of point charge from second charge

Using formula for distance

$d=16-x$

Put the value of x

$d=16-4$

$d=12\ cm$

Hence, The net electrostatic force will be zero at 4 cm from first charge and 12 cm from second charge.