Question

Two charges $ q_1 $ and $q_2$ are placed 30 cm apart, as shown in the figure.

A third charge $q_3$ is moved along the are of a circle of radius 40cm from C to D. The change in the potential energy of the system is $ \frac{q_3}{ 4 \pi \varepsilon_0}$ k. where k is

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

Answer 1

Answer: (C)

The potential energy when $q_3$ is at point C
$ U_1 = \frac{1}{ 4 \pi \varepsilon_0} \bigg [ \frac{ q_1 q_3}{ 0.40} + \frac{ q_2 q_3}{ \sqrt{ (0.40)^2 + (0 . 30)^2 }} \bigg ] $
The potential energy when $q_3$ is at point D
$ U_2 = \frac{1}{ 4 \pi \varepsilon_0} \bigg [ \frac{ q_1 q_3}{ 0.40} + \frac{q_2 q_3 }{ 0.10} \bigg ] $
Thus change in potential energy is

$ \triangle U = U_2 - U_1$
$ \Rightarrow \frac{q_3}{ 4 \pi \varepsilon_0} = k $
= $ \frac{1}{ 4 \pi \varepsilon_0} \bigg [ \frac{ q_1 q_3}{ 0.40} + \frac{ q_2 q_3 }{ 0.10} - \frac{ q_1 q_3}{ 0 . 40} - \frac{ q_2 q_3 }{ 0 . 50 } \bigg ] $
$\Rightarrow k = \frac{ 5 q_2 - q_2 }{ 0 . 50 } = \frac{ 4q_2}{ 0 . 50} = 8 q_2$.