A spherical shell of radius r with a uniform charge q has point charge `q_(0)` at its centre. Find the work performed by the electric forces during the shell expansion slowly from radius R to 2R. Also find out work done by external agent against electric forces.

Answer

To solve the problem, we need to calculate the work done by electric forces during the expansion of a spherical shell with a uniform charge \( q \) from radius \( R \) to \( 2R \) while a point charge \( q_0 \) is placed at its center. We will also find the work done by an external agent against electric forces.

### Step 1: Calculate Initial Potential Energy \( U_i \)

The initial potential energy \( U_i \) consists of two parts: the self-energy of the spherical shell and the interaction energy between the shell and the point charge.

1. **Self-energy of the spherical shell**:
\[ U_{self} = \frac{kq^2}{2R} \]
where \( k \) is Coulomb's constant.

2. **Interaction energy between the shell and the point charge**:
\[ U_{interaction} = \frac{kqq_0}{R} \]

Thus, the initial potential energy \( U_i \) is:
\[ U_i = U_{self} + U_{interaction} = \frac{kq^2}{2R} + \frac{kqq_0}{R} \]

### Step 2: Calculate Final Potential Energy \( U_f \)

When the radius expands to \( 2R \), we calculate the final potential energy \( U_f \):

1. **Self-energy of the spherical shell**:
\[ U_{self} = \frac{kq^2}{2(2R)} = \frac{kq^2}{4R} \]

2. **Interaction energy between the shell and the point charge**:
\[ U_{interaction} = \frac{kqq_0}{2R} \]

Thus, the final potential energy \( U_f \) is:
\[ U_f = U_{self} + U_{interaction} = \frac{kq^2}{4R} + \frac{kqq_0}{2R} \]

### Step 3: Calculate Change in Potential Energy \( \Delta U \)

The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]

Substituting the expressions for \( U_f \) and \( U_i \):
\[ \Delta U = \left(\frac{kq^2}{4R} + \frac{kqq_0}{2R}\right) - \left(\frac{kq^2}{2R} + \frac{kqq_0}{R}\right) \]

### Step 4: Simplify \( \Delta U \)

Now simplifying \( \Delta U \):
\[ \Delta U = \frac{kq^2}{4R} + \frac{kqq_0}{2R} - \frac{kq^2}{2R} - \frac{kqq_0}{R} \]
\[ = \frac{kq^2}{4R} - \frac{2kq^2}{4R} + \frac{2kqq_0}{4R} - \frac{4kqq_0}{4R} \]
\[ = -\frac{kq^2}{4R} - \frac{2kqq_0}{4R} \]
\[ = -\frac{k}{4R}(q^2 + 2qq_0) \]

### Step 5: Work Done by External Agent

The work done by the external agent \( W_{ext} \) is equal to the change in potential energy:
\[ W_{ext} = \Delta U = -\frac{k}{4R}(q^2 + 2qq_0) \]

### Step 6: Work Done by Electric Forces

The work done by electric forces \( W_{el} \) is equal in magnitude but opposite in sign to the work done by the external agent:
\[ W_{el} = -W_{ext} = \frac{k}{4R}(q^2 + 2qq_0) \]

### Final Answers

1. Work done by the external agent:
\[ W_{ext} = -\frac{k}{4R}(q^2 + 2qq_0) \]

2. Work done by electric forces:
\[ W_{el} = \frac{k}{4R}(q^2 + 2qq_0) \]