Consider a spherical shell of radius R wil a total charge +Q uniformly spread on its surface (centre of the shell lies at the origin x =0) Two point charge +q and -q are brought, one after the other, from far away and placed at `x = -a//2` and `x = +a//2(a lt R)` respectively. Magnitude of the work done in this process is

Options

A. a. `(Q+q)^(2)//4 pi epsilon_(0)a`
B. b. `zero`
C. c. `q^(2)//4 pi epsilon_(0)a`
D. d. `Qq//4 pi epsilon_(0)a`

Explanation

To solve the problem, we need to calculate the work done in bringing two point charges (+q and -q) from infinity to specific positions near a uniformly charged spherical shell. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have a spherical shell of radius \( R \) with a total charge \( +Q \) uniformly spread on its surface. The center of the shell is at the origin (0,0,0). We need to bring a charge \( +q \) to the position \( x = -\frac{a}{2} \) and a charge \( -q \) to the position \( x = +\frac{a}{2} \), where \( a < R \).

### Step 2: Calculate the Work Done to Bring Charge +q
The work done \( W_1 \) in bringing the charge \( +q \) from infinity to the point \( x = -\frac{a}{2} \) can be calculated using the formula:
\[ W_1 = q \cdot V \]
where \( V \) is the electric potential at the point \( x = -\frac{a}{2} \).

For a uniformly charged spherical shell, the electric potential inside the shell (at any point within the radius \( R \)) is given by:
\[ V = \frac{kQ}{R} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

Thus, the work done in bringing the charge \( +q \) is:
\[ W_1 = +q \cdot \frac{kQ}{R} \]

### Step 3: Calculate the Work Done to Bring Charge -q
Next, we bring the charge \( -q \) to the position \( x = +\frac{a}{2} \). The potential at this point is still the same as at the previous point because it is also inside the shell:
\[ V' = \frac{kQ}{R} \]

The work done \( W_2 \) in bringing the charge \( -q \) is:
\[ W_2 = -q \cdot V' = -q \cdot \frac{kQ}{R} \]

### Step 4: Calculate the Total Work Done
The total work done \( W \) in bringing both charges is the sum of the work done for each charge:
\[ W = W_1 + W_2 \]
Substituting the values we found:
\[ W = \left( +q \cdot \frac{kQ}{R} \right) + \left( -q \cdot \frac{kQ}{R} \right) \]
This simplifies to:
\[ W = \frac{kQq}{R} - \frac{kQq}{R} = 0 \]

### Step 5: Magnitude of Work Done
Since the total work done is zero, the magnitude of the work done is:
\[ |W| = 0 \]

### Final Answer
The magnitude of the work done in this process is \( 0 \).

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