Question

A rod is bent into a semi-circular arc of radius R. The rod has a uniform linear charge distribution $\lambda$. The potential at the centre of the arc, point P is

• A. $\frac{\lambda }{2\pi {ϵ}_{0}R}$
• B. $\frac{\lambda }{4{ϵ}_0}$
• C. $\frac{\lambda }{2{ϵ}_0}$
• D. $\frac{\lambda }{{ϵ}_0}$

Answer 1

The correct option is B $\frac{\lambda }{4{ϵ}_0}$

Now,

$\lambda =\frac{Q}{L}$

$\lambda =\frac{Q}{\pi R}$

$dq=\lambda dx$

$=\frac{Q}{\pi R}dx$

Potential at $P$

$\int d{V}_1=\int \frac{Kdq}{R}$

$=\int \frac{KQ}{R\pi }\frac{dx}{R}$

$={\int }_0^{\pi R}\frac{Q}{\pi R}\times \frac{dx}{4\pi {\epsilon }_{0}R}$

$V_p=\frac{Q}{4\pi {\epsilon }_{0}R}$

$V_p=\frac{\lambda \times \pi R}{4\pi {\epsilon }_{0}R}$

$V_p=\frac{\lambda }{4{\epsilon }_0}$

Hence option (B) is correct