Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 ….. ∞. Here x0 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/4πε0r. Then the potential at the origin due to the above system of charges is :

Options:

0

$\frac{q}{8 \pi \varepsilon_0 x_0 \ln 2}$

$-\frac{q \ln (2)}{4 \pi \varepsilon_0 x_0}$

Correct Answer:

$-\frac{q \ln (2)}{4 \pi \varepsilon_0 x_0}$

Explanation:

Potential at origin will be given by

V = $\frac{q}{4 \pi \rho_0}\left[\frac{1}{x_0}-\frac{1}{2 x_0}+\frac{1}{3 x_0}-\frac{1}{4} x_0+.....\right]$

$=\frac{q}{4 \pi \varepsilon_0} . \frac{1}{x_0}\left[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....\right]$

$=\frac{q}{4 \pi \varepsilon_0 x_0} \ln (2)$