Twosmall neutral conducting spheres are taken and their centres are separated by a distance 2a. If n electrons are removed from one of them and deposited on the other, what is the magnitude of electric intensity due to the system at a point on the line joining the centres of the spheres and at a distance d from the midpoint of the line joining the two spheres? (e is quantum of charge)

Options

A. `(2 "ne"a)/(2p e_(0)d^(3))`
B. `(2n^(2)e^(2)a)/(2pe_(0)d^(3))`
C. `(4"ne" a)/(4p e_(0)d^(3))`
D. `(4n^(2)e^(2)a)/(4pe_(0)d^(3))`

Explanation

To solve the problem, we need to find the magnitude of the electric field intensity at a point on the line joining the centers of two small neutral conducting spheres, after transferring \( n \) electrons from one sphere to the other.

### Step-by-Step Solution:

1. **Understanding the Charge Transfer**:
- Initially, both spheres are neutral. When \( n \) electrons are removed from one sphere and deposited on the other, the sphere from which electrons are removed will have a positive charge, and the sphere that gains electrons will have a negative charge.
- The charge of one electron is denoted by \( e \). Thus, the charge transferred is:
\[ q = n \cdot e \]
- The sphere losing electrons will have a charge of:
\[ +q = +n \cdot e \]
- The sphere gaining electrons will have a charge of:
\[ -q = -n \cdot e \]

2. **Setting Up the Geometry**:
- Let the distance between the centers of the two spheres be \( 2a \).
- The midpoint between the two spheres is at a distance \( a \) from each sphere.
- We need to find the electric field at a point \( P \) which is at a distance \( d \) from the midpoint.

3. **Calculating Distances**:
- The distance from the positive charge \( +q \) (first sphere) to point \( P \) is:
\[ r_1 = d - a \]
- The distance from the negative charge \( -q \) (second sphere) to point \( P \) is:
\[ r_2 = d + a \]

4. **Calculating Electric Fields**:
- The electric field \( E_1 \) due to the positive charge \( +q \) at point \( P \) is given by:
\[ E_1 = \frac{k \cdot q}{(d - a)^2} \]
- The electric field \( E_2 \) due to the negative charge \( -q \) at point \( P \) is given by:
\[ E_2 = \frac{k \cdot q}{(d + a)^2} \]
- The direction of \( E_1 \) is away from the positive charge, while the direction of \( E_2 \) is towards the negative charge.

5. **Net Electric Field**:
- The net electric field \( E \) at point \( P \) is the difference between \( E_1 \) and \( E_2 \):
\[ E = E_1 - E_2 = \frac{k \cdot q}{(d - a)^2} - \frac{k \cdot q}{(d + a)^2} \]

6. **Simplifying the Expression**:
- Factor out \( k \cdot q \):
\[ E = k \cdot q \left( \frac{1}{(d - a)^2} - \frac{1}{(d + a)^2} \right) \]
- To combine the fractions, use a common denominator:
\[ E = k \cdot q \cdot \frac{(d + a)^2 - (d - a)^2}{(d - a)^2 (d + a)^2} \]
- Expanding the numerator:
\[ (d + a)^2 - (d - a)^2 = (d^2 + 2da + a^2) - (d^2 - 2da + a^2) = 4da \]
- Therefore, we have:
\[ E = k \cdot q \cdot \frac{4da}{(d - a)^2 (d + a)^2} \]

7. **Substituting for \( q \)**:
- Substitute \( q = n \cdot e \):
\[ E = k \cdot n \cdot e \cdot \frac{4da}{(d - a)^2 (d + a)^2} \]

8. **Final Expression**:
- The final expression for the electric field intensity at point \( P \) is:
\[ E = \frac{4k n e da}{(d - a)^2 (d + a)^2} \]