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. (d)
Solution:
- Potential at origin will be given by
$$ \begin{aligned} V & =\frac{q}{4 \pi \varepsilon_{0}} \Big[\frac{1}{x_{0}}-\frac{1}{2 x_{0}}+\frac{1}{3 x_{0}}-\frac{1}{4 x_{0}}+\ldots .\Big] \\ & =\frac{q}{4 \pi \varepsilon_{0}} \cdot \frac{1}{x_{0}} \Big[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots\Big]=\frac{q}{4 \pi \varepsilon_{0} x_{0}} \ln (2) \end{aligned} $$
