Question

Two charges $q_1$ and $q_2$ are placed $30cm$ apart as shown. A third charge $q_3$ is moved along the circle of radius $40cm$ from $C$ to $D$. The change in the potential energy of the system is $\frac{q_{3}K}{4\pi {ϵ}_0}$

Find K?

• A. $8q_2$
• B. $8q_1$
• C. $6q_2$
• D. $6q_1$

Answer 1

The correct option is A $8q_2$

Given two charges $q_1$ and $q_2$ are placed $30cm$ apart as shown inn the figure. A third charge $q_3$ moved from $C$ to $D$.

Radius $=40cm$

Given the change in potential energy of the system is $\frac{q_{3}k}{4\pi {\epsilon }_0}$. We have to find the value of $k$.

We solve it as follows:

The change in potential energy of system is $U_D-U_C$ as discussed under:

When charge $q_3$ is at $C$, then its potential energy is $U_C=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1{q}_3}{0.4}+\frac{q_2{q}_3}{0.5}]$

when charge $q_3$ is at $D$, then $U_D$ is

$U_D=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1{q}_3}{0.4}+\frac{q_2{q}_3}{0.1}]$

Hence, change in potential energy is $\Delta U$

$\Delta U=\frac{1}{4\pi {\epsilon }_0}[\frac{q_2{q}_3}{0.1}-\frac{q_2{q}_3}{0.5}]$ ...(i)

But given, $\Delta U=\frac{q_{3}k}{4\pi {\epsilon }_0}$ ...(ii)

On equating (i) and (ii), we get

$\frac{q_{3}k}{4\pi {\epsilon }_0}=\frac{1}{4\pi {\epsilon }_0}[\frac{q_2{q}_3}{0.1}-\frac{q_2{q}_3}{0.5}]$

We get,

$k=q_2(10-2)=8q_2$