Electrostatics 2 Question 14
15. A charge $+q$ is fixed at each of the points $x=x _0, x=3 x _0, x=5 x _0 \ldots \infty$ on the $x$-axis and a charge $-q$ is fixed at each of the points $x=2 x _0, x=4 x _0, x=6 x _0 \ldots, \infty$. Here, $x _0$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q / 4 \pi \varepsilon _0 r$. Then the potential at the origin due to the above system of charges is
(1998, 2M)
(a) zero
(b) $\frac{q}{8 \pi \varepsilon _0 x _0 \ln 2}$
(c) infinite
(d) $\frac{q \ln (2)}{4 \pi \varepsilon _0 x _0}$
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Answer:
Correct Answer: 15. (a, c)
Solution:
- Potential at origin will be given by
$$ \begin{aligned} V & =\frac{q}{4 \pi \varepsilon _0} \frac{1}{x _0}-\frac{1}{2 x _0}+\frac{1}{3 x _0}-\frac{1}{4 x _0}+\ldots . \\ & =\frac{q}{4 \pi \varepsilon _0} \cdot \frac{1}{x _0} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots=\frac{q}{4 \pi \varepsilon _0 x _0} \ln (2) \end{aligned} $$
