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Q.

Two charges $q_{1} and q_{2}$ are placed 30 cm apart, as shown in Fig. 2.135. A third charge $q_{3}$ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy I of the system is $(\frac{q_{3}}{4 {πε}_{0}}) k$ , where k is :

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a

$6 q_{1}$

b

$8 q_{2}$

c

$6 q_{2}$

d

$8 q_{1}$

answer is A.

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Detailed Solution

$U_{1} = \frac{1}{4 {πε}_{0}} (\frac{q_{1} q_{3}}{(0.4)} + \frac{q_{1} q 2}{(0.3)} + \frac{q_{2} q_{3}}{(0.5)}) U_{2} = \frac{1}{4 {πε}_{0}} (\frac{q_{1} q_{3}}{(0.4)} + \frac{q_{1} q 2}{(0.3)} + \frac{q_{2} q_{3}}{(0.1)})$

$\Rightarrow ∆ U = U_{2} - U_{1} = \frac{1}{4 {πε}_{0}} q_{2} q_{3} (\frac{1}{0.1} - \frac{1}{0.5}) = \frac{q_{3}}{4 {πε}_{0}} (8 q_{2}) \Rightarrow k = 8 q_{2}$

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