Electrostatics 2 Question 10
10. A charge $Q$ is uniformly distributed over a long $\operatorname{rod} A B$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is
(2013 Main)
(a) $\frac{Q}{8 \pi \varepsilon_{0} L}$
(b) $\frac{3 Q}{4 \pi \varepsilon_{0} L}$
(c) $\frac{Q}{4 \pi \varepsilon_{0} L \ln 2}$
(d) $\frac{Q \ln 2}{4 \pi \varepsilon_{0} L}$
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Answer:
Correct Answer: 10. (d)
Solution:
$$ \begin{aligned} V=\int_ {L}^{2 L} \frac{k d Q}{x} & =\int_ {L}^{2 L} \frac{k \frac{Q}{L} d x}{x}=\frac{Q}{4 \pi \varepsilon_ {0} L} \int_ {L}^{2 L} \frac{1}{x} d x \\ & =\frac{Q}{4 \pi \varepsilon_ {0} L}\left[\log _ {e} x\right]_ {L}^{2 L} \\ & =\frac{Q}{4 \pi \varepsilon_ {0} L}\left[\log _ {e} 2 L-\log _ {e} L\right]=\frac{Q}{4 \pi \varepsilon_{0} L} \ln (2) \end{aligned} $$
