Question

The following diagram represents a semi circular wire of linear charge density $\lambda ={\lambda }_0\sin \theta$, where ${\lambda }_0$ is a positive constant. The electric potential at $\text{O}$ is $(\text{take}k=\frac{1}{4\pi {ϵ}_0})$

• A. $k{\lambda }_0\sin \theta$
• B. $\frac{k{\lambda }_0\cos \theta }{2}$
• C. Zero
• D. $k{\lambda }_0\cos \theta$

Answer 1

The correct option is C Zero

Given,

linear charge density, $\lambda ={\lambda }_0\sin \theta$

Let us consider a small element of length $dl$ on wire.

So, charge on the element, $dQ$ can be written as,

$dQ=\lambda \left(dl\right)$

From the data given in the question and from the figure we can deduce that,

$dQ={\lambda }_0\sin \theta R\left(d\theta \right)$

Potential at the centre $O$ due to this element
$dV=\frac{1}{4\pi {ϵ}_0}\frac{{\lambda }_{0}R\sin \theta \left(d\theta \right)}{R}$

Now the potential at $O$ due to whole arc, can be calculated by integrating it,

$V=\int dV=\frac{1}{4\pi {ϵ}_0}{\lambda }_0\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\sin \theta \left(d\theta \right)$

$\Rightarrow V=-\frac{{\lambda }_0}{4\pi {ϵ}_0}(\cos \theta {)}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$\therefore V=0$

Hence option (c) is the correct answer.