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. (c)
Solution:
$$ \begin{aligned} & \\ V=\int _L^{2 L} \frac{k d Q}{x} & =\int _L^{2 L} \frac{k \frac{Q}{L} d x}{x}=\frac{Q}{4 \pi \varepsilon _0 L} \int _L^{2 L} \frac{1}{x} d x \\ & =\frac{Q}{4 \pi \varepsilon _0 L}\left[\log _e x\right] _L^{2 L} \\ & =\frac{Q}{4 \pi \varepsilon _0 L}\left[\log _e 2 L-\log _e L\right]=\frac{Q}{4 \pi \varepsilon _0 L} \ln (2) \end{aligned} $$
11. For inside points $(r \leq R)$
$E=0 \Rightarrow V=$ constant $=\frac{1}{4 \pi \varepsilon _0} \frac{q}{R}$
For outside points $(r \geq R)$
$$ \text { and } \begin{aligned} E & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{q}{r^{2}} \quad \text { or } \quad E & \propto \frac{1}{r^{2}} \\ \text { an } & =\frac{1}{4 \pi \varepsilon _0} \frac{q}{r} \quad \text { or } \quad V & \propto \frac{1}{r} \end{aligned} $$
On the surface $(r=R)$
$$ \begin{aligned} V & =\frac{1}{4 \pi \varepsilon _0} \frac{q}{R} \\ \Rightarrow \quad E & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{q}{R^{2}}=\frac{\sigma}{\varepsilon _0} \end{aligned} $$
where, $\sigma=\frac{q}{4 \pi R^{2}}=$ surface charge density corresponding to above equations the correct graphs are shown in option (d).
