Question

A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.57 m. A third charge of +2q is located on the x axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

Answer 1

Answer:

The third charge placed is 0.80 m.

Explanation:

Given that,

Distance = 0.57 m

First charge = q

Third charge = 2q

We need to calculate the electrostatic force on charge q₁ due to q₂

Using formula of electrostatic force

$F_{21}=\dfrac{kq_{1}q_{2}}{r_{1}^2}$

When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is

$F_{net}=F_{21}+F_{31}$

The net electrostatic force on the charge at the origin doubles.

$2F_{21}=F_{21}+F_{31}$

$F_{31}=F_{21}$

$\dfrac{kq_{3}q_{1}}{r_{2}^2}=\dfrac{kq_{2}q_{1}}{r_{1}^2}$

$\dfrac{q_{3}}{r_{2}^2}=\dfrac{q_{2}}{r_{1}^2}$

$r_{2}^2=\dfrac{q_{3}}{q_{2}\timesr_{1}^2}$

$r_{2}=\sqrt{\dfrac{q_{3}}{q_{2}}}r_{1}$

Put the value into the formula

$r_{2}=\sqrt{\dfrac{2q}{q}}\times0.57$

$r_{2}=\sqrt{2}\times0.57$

$r_{2}=0.80\ m$

Hence, The third charge placed is 0.80 m from origin in x-axis.