Question

Calculate the electric potential at point $R$ on the $x-$axis.

• A. $0$
• B. $V=\frac{q}{4\pi {ϵ}_o\sqrt{a^2+y^2}}$
• C. $V=\frac{q}{4\pi {ϵ}_o}(\frac{1}{x+a}-\frac{1}{x-a})$
• D. $V=\frac{q}{4\pi {ϵ}_o}(\frac{1}{x+a}+\frac{1}{x-a})$

Answer 1

The correct option is C $V=\frac{q}{4\pi {ϵ}_o}(\frac{1}{x+a}-\frac{1}{x-a})$

Potential at any point R $V=\frac{q}{4\pi {ϵ}_{o}r}$

where $r$ is the distance of the point $R$ and the charge.

Thus potential at a point $R$ at position $x$, $V=V_q+V_{-q}$

$\therefore$ $V=\frac{q}{4\pi {ϵ}_o(x+a)}+\frac{-q}{4\pi {ϵ}_o(x-a)}$

$⟹$ $V=\frac{q}{4\pi {ϵ}_o}(\frac{1}{x+a}-\frac{1}{x-a})$