Question

A Charged Configuration is As Shown In The Figure. What Is The potential At a Point P At a Distance r On The Axis The Axis The Shown assuming r>> a?

• A. $\frac{q}{4\pi E_o}(\frac{2a}{r^2}+\frac{1}{r})$
• B. $\frac{q}{4\pi E_o}(\frac{2a}{r}+\frac{1}{r^2})$
• C. $\frac{q}{4\pi E_o}(\frac{a}{r^2}+\frac{1}{r})$
• D. $\frac{q}{4\pi E_o}(\frac{a}{r^2}+\frac{2}{r})$

Answer 1

The correct option is A $\frac{q}{4\pi E_o}(\frac{2a}{r^2}+\frac{1}{r})$

Potential at P

$\Rightarrow V_p=\frac{1}{4\pi E_0}[\frac{q}{r-a}+\frac{q}{r}-\frac{q}{r+a}]$

$=\frac{q}{4\pi E_0}[\frac{(r+a)r+a}{r^2-a^2}+\frac{1}{r}]$ since $r>>a,r^2-a^2=r^2$

$=\frac{q}{4\pi E_0}[\frac{2a}{r^2-a^2}+\frac{1}{r}]$

$=\frac{q}{4\pi E_0}[\frac{2a}{r^2}+\frac{1}{r}]$

Hence, the answer is $\frac{q}{4\pi E_0}[\frac{2a}{r^2}+\frac{1}{r}].$