Question

A charge $+q$ is fixed at each of the point $x=x_0,x=3x_0,x=5x_0...$ upto infinity and a charge $-q$ fixed at each of the points $x=2x_0,x=4x_0,x=6x_0...$ upto infinity. Here $x_0$ is a positive constant. The potential at the origin due to this system of charges is

• A. Infinity
• B. Zero
• C. $\frac{q}{4\pi {\epsilon }_0{x}_0{\log }_{e\left(2\right)}}$
• D. $\frac{q{\log }_e\left(2\right)}{4\pi {\epsilon }_0{x}_0}$

Answer 1

The correct option is D $\frac{q{\log }_e\left(2\right)}{4\pi {\epsilon }_0{x}_0}$

Potential due to $+q$ charges,

$V_1=\frac{1}{4\pi {\epsilon }_0}\{\frac{q}{x_0}+\frac{q}{3x_0}+\frac{q}{5x_0}+....\infty \}$

Potential due to $-q$ charges,

$V_2=\frac{1}{4\pi {\epsilon }_0}\{\frac{-q}{2x_0}+\frac{-q}{{4x}_0}+\frac{-q}{6x_0}+...\infty \}$

Total potential at the origin,

$V=V_1+V_2$

$\Rightarrow V=\frac{1}{4\pi {\epsilon }_0}\{\frac{q}{x_0}+\frac{q}{3x_0}+\frac{q}{5x_0}+....\infty \}+\frac{1}{4\pi {\epsilon }_0}\{\frac{-q}{2x_0}+\frac{-q}{x_0}+\frac{-q}{6x_0}+...\infty \}$

$\Rightarrow V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{x_0}\{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...\infty \}$

$\left[{\log }_e\right(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}......\infty ]$

$\Rightarrow V=\frac{q}{4\pi {\epsilon }_0{x}_0}{\log }_e(1+1)$

$\Rightarrow V=\frac{q{\log }_e\left(2\right)}{4\pi {\epsilon }_0{x}_0}$

Hence, option (d) is the correct answer.