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: Understanding the Question:
We need to determine the total electrostatic potential ($V$) at the geometric center of a semicircular half-ring of radius $R$ that carries a uniform linear charge density $\sigma$.

Step 2: Detailed Explanation:
The total charge $q$ distributed uniformly across the half-ring is equal to the product of its linear charge density and its total arc length ($\pi R$): $$ q = \sigma \times \pi R $$ Since every single infinitesimal charge element $dq$ on the ring sits at the exact same radial distance $R$ away from the center point, the electric potential is found by straightforward scalar accumulation: $$ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{R} $$ Substituting our expression for the total charge $q$: $$ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{\sigma \cdot \pi R}{R} $$ Canceling out the common $\pi$ and radius $R$ parameters from the numerator and denominator: $$ V = \frac{\sigma}{4\epsilon_0} $$

Step 3: Final Answer:
The electric potential at the center is $\frac{\sigma}{4\epsilon_0}$, which corresponds to option (D).