Electrostatic Potential and Capacitance - Result Question 16
18. As per the diagram, a point charge $+q$ is placed at the origin $O$. Work done in taking another point charge $-Q$ from the point $A$ [coordinates $(0, a)]$ to another point $B$ [coordinates $(a, 0)]$ along the straight path $A B$ is:
[2005]
(a) zero
(b) $(\frac{-q Q}{4 \pi \varepsilon_0} \frac{1}{a^{2}}) \sqrt{2} a$
(c) $(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^{2}}) \cdot \frac{a}{\sqrt{2}}$
(d) $(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^{2}}) \cdot \sqrt{2} a$
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Answer:
Correct Answer: 18. (a)
Solution:
- (a) We know that potential energy of two charge system is given by
$U=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}$ According to question,
$U_A=\frac{1}{4 \pi \epsilon_0} \frac{(+q)(-Q)}{a}=-\frac{1}{4 \pi \varepsilon_0} \frac{Q q}{a}$
and $U_B=\frac{1}{4 \pi \epsilon_0} \frac{(+q)(-Q)}{a}=-\frac{1}{4 \pi \varepsilon_0} \frac{Q q}{a}$
$\Delta U=U_B-U_A=0$
We know that for conservative force, $W=-\Delta U=0$
