Q. 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 arc 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

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Solution:

The potential energy when q3 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 ] $
$\Rightarrow k = \frac{ 5 q_2 - q_2 }{ 0 . 50 } = \frac{ 4q_2}{ 0 . 50} = 8 q_2$

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