Electrostatics Ques 163
- Two equal point charges are fixed at $x=-a$ and $x=+a$ on the $x$-axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$, when it is displaced by a small distance $x$ along the $x$-axis, is approximately proportional to
$(2002,1 \mathrm{M})$
(a) $x$
(b) $x^{2}$
(c) $x^{3}$
(d) $1 / x$
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Answer:
Correct Answer: 163.(b)
Solution:
Formula:
Coulomb Force Between Two Point Charges:
$$ \begin{aligned} U_{i} & =\frac{2 K Q q}{a}+\frac{K \cdot q \cdot q .}{2 a} \\ \text { and } \quad U_{f} & =K Q q \Big[\frac{1}{a+x}+\frac{1}{a-x}\Big]+\frac{K \cdot q \cdot q .}{2 a} \end{aligned} $$
$$ \begin{aligned} & \text { Here, } \quad K=\frac{1}{4 \pi \varepsilon_{0}} \\ & \Delta U=U_{f}-U_{i} \\ & \text { or } \quad|\Delta U|=\frac{2 K Q q x^{2}}{a^{3}} \\ & \text { For } \quad x«a \\ & \therefore \quad \Delta U \propto x^{2} \end{aligned} $$
BETA
