Problem 39 - A Complete Resource Book in Physics for JEE Main

Question

A half ring of radius \(R\) has a charge of \(\lambda\) per unit length. The potential at the centre of the half ring is (A) \(k \frac{\lambda}{R}\) (B) \(k \frac{\lambda}{\pi R}\) (C) \(k \frac{\pi \lambda}{R}\) (D) \(k \pi \lambda\)

Step-by-Step Solution

Answer

The short answer is: (C) \(k \cfrac{\pi \lambda}{R}\).

Step 1: Set up the integral for electric potential

Firstly, let's break the half ring into tiny charge elements (dQ) and sum up the electric potential (dV) at the center of the half ring due to each charge element. To get the electric potential (V) at the center, we integrate the electric potential (dV) due to the tiny charge elements (dQ) around the half ring. For a given point dQ on the half ring, the electric potential dV at the center can be given by: \(dV = \cfrac{k dQ}{r}\), where r is the distance between the point charge dQ and the center of the half-ring. Since the distance between any point on the half-ring and the center is always R, we can write it as \(dV = \cfrac{k dQ}{R}\) Now, we should express dQ in terms of λ and integrate over the entire half ring.

Step 2: Express the charge element (dQ) in terms of λ

We know that λ is the charge per unit length. So, the charge dQ in a tiny arc length dl can be given as: \(dQ = \lambda dl \) To integrate around the half ring, we can introduce an angle dθ and relate it to the arc length dl by \(dl = R dθ\) Now substituting the arc length into the expression for dQ, we get \(dQ = \lambda R dθ\) Now we should substitute this expression for dQ into the electric potential expression and integrate over the entire half ring.

Step 3: Integrate the electric potential expression

Now substituting the expression for dQ into the electric potential expression, \(dV = \cfrac{k (\lambda R dθ)}{R}\) Integrating the expression for dV with respect to θ over the range from 0 to π (since it is a half-ring), \(V = \int_{0}^{\pi} \cfrac{k \lambda R dθ}{R}\) Which simplifies to \(V = \cfrac{k \lambda}{R} \int_{0}^{\pi} dθ\) Now, integrating with respect to θ, we get \(V = \cfrac{k \lambda}{R} (\pi - 0)\) So, the expression for the electric potential at the center of the half-ring is \(V = \cfrac{k \pi \lambda}{R}\) Looking at the options given in the exercise, we see that the expression we obtained matches option (C). Therefore, the correct answer is (C) \(k \cfrac{\pi \lambda}{R}\).