Question

A uniformly charged half ring of a radius 'R' has linear charge density '$\sigma$'. The electric potential at the centre of the half ring is ($\epsilon_0$ = permittivity of free space)

• A. $\frac{\sigma}{6\epsilon_0}$
• B. $\frac{\sigma}{2\epsilon_0}$
• C. $\frac{\sigma}{\epsilon_0}$
• D. $\frac{\sigma}{4\epsilon_0}$

Hint:

Because electric potential is a scalar quantity, the geometric shape or folding of the wire doesn't matter! As long as all charges stay at distance $R$, the potential formula remains simply $\frac{1}{4\pi\epsilon_0} \cdot \frac{\text{Total Charge}}{R}$. Substituting $\text{Charge} = \sigma \pi R$ clears the terms instantly!

Answer 1

The Correct Option is D

Step 1: Find the total charge.
The half ring has length $\pi R$, so its charge is $q=\sigma\cdot\pi R$.

Step 2: Use equal distance for potential.
Every bit of charge is the same distance $R$ from the centre, so the potential adds up simply as $V=\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{R}$.

Step 3: Substitute and cancel.
\[ V=\frac{1}{4\pi\epsilon_0}\cdot\frac{\sigma\pi R}{R}=\frac{\sigma}{4\epsilon_0} \]
\[ \boxed{\frac{\sigma}{4\epsilon_0}} \]