Work done by an external agent to move slowly a charge Q from rim of a uniformly charged horizontal disc of radius R and charge per unit area `sigma`, to center of this disc is

Options

A. `(sigmaRQ)/(epsilon_(0))((2)/(pi)-(1)/(2))`
B. `(sigmaRQ)/(epsilon_(0))((1)/(2)-(1)/(pi))`
C. `(sigmaaQ)/(epsilon_(0))((1)/(pi)-(1)/(2))`
D. `(sigmaaQ)/(epsilon_(0))((1)/(2)-(2)/(pi))`

Answer

To find the work done by an external agent to move a charge \( Q \) from the rim of a uniformly charged horizontal disc of radius \( R \) and charge per unit area \( \sigma \) to the center of the disc, we will follow these steps:

### Step 1: Understand the Potential at the Center and Rim of the Disc
1. **Potential at the Center of the Disc**:
The potential \( V_c \) at the center of a uniformly charged disc is given by the formula:
\[ V_c = \frac{\sigma R}{2 \epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

2. **Potential at the Rim of the Disc**:
The potential \( V_r \) at the rim (edge) of the disc is given by:
\[ V_r = \frac{\sigma R}{\pi \epsilon_0} \]

### Step 2: Calculate the Potential Difference
The work done \( W \) by the external agent in moving the charge \( Q \) from the rim to the center is given by the formula:
\[ W = Q \cdot (V_c - V_r) \]
Substituting the values of \( V_c \) and \( V_r \):
\[ W = Q \left( \frac{\sigma R}{2 \epsilon_0} - \frac{\sigma R}{\pi \epsilon_0} \right) \]

### Step 3: Simplify the Expression
Factor out common terms:
\[ W = \frac{Q \sigma R}{\epsilon_0} \left( \frac{1}{2} - \frac{1}{\pi} \right) \]

### Step 4: Final Expression for Work Done
Thus, the work done by the external agent to move the charge \( Q \) from the rim to the center of the disc is:
\[ W = \frac{Q \sigma R}{\epsilon_0} \left( \frac{1}{2} - \frac{1}{\pi} \right) \]