Question

A charge $+q$ is fixed at each of the points $x=x_0,x=3x_0,x=5x_0.....,\infty$ on the x-axis and charge $-q$ is fixed at each of the points $x=3x_0,x=4x_0,x=6x_0,.....,\infty$. Here, $x_0$ is a positive quantity. Take the electric potential at a point due to charge $Q$ at a distance $r$ from it to be $\frac{Q}{4\pi {\epsilon }_{0}r}$. Then the potential at origin due to the above system of charges is:

• A. zero
• B. Infinity
• C. $\frac{q}{8\pi {\epsilon }_0{x}_{0}In2}$
• D. $\frac{qIn2}{4\pi {\epsilon }_0{x}_0}$

Answer 1

The correct option is D $\frac{qIn2}{4\pi {\epsilon }_0{x}_0}$

Potential at origin due to positive charge

$V(+)=\frac{Kq}{x_0}+\frac{Kq}{3x_0}+\frac{Kq}{5x_0}......\infty$

Similarly,

$V(-)=-[\frac{Kq}{{2x}_0}+\frac{Kq}{4x_0}+\frac{Kq}{6x_0}......\infty ]$

Net potential

$V_{net}=\frac{Kq}{x_0}-\frac{Kq}{{2x}_0}+\frac{Kq}{3x_0}-\frac{Kq}{{4x}_0}......\infty$

$\Rightarrow V_{net}=\frac{Kq}{x_0}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}......\infty ]$

$\Rightarrow V_{net}=\frac{Kq}{x_0}ln2$