You are moving a negative charge `q lt 0` at a small constant speed away from a uniformly charged non-conducting spherical shell on which resides a negative charge `Q lt 0`. The electrostatic, field of `Q` is `E`. Let `U` be the total energy of the system, `W_(a)` the work done by the force `F_(a)` you exert on `q, W_(E)` the work done by the electrostaitc force `F_(E)` on `q`. Then, as `q` is being moved :

Options

A. `W_(a) = -W_(E)`, there `U` remains constant
B. `F_(a) = -F_(E)`
C. `U` increases
D. `U` decreases

Answer

To solve the problem step-by-step, we need to analyze the forces acting on the negative charge \( q \) as it is moved away from the uniformly charged non-conducting spherical shell with charge \( Q \).

### Step 1: Understanding the Forces
When moving a negative charge \( q \) away from a negatively charged shell, the electrostatic force \( F_E \) acting on \( q \) due to the shell is repulsive because both charges are negative. The force exerted by you, \( F_A \), must also act in the direction opposite to \( F_E \) to maintain a constant speed.

**Hint:** Remember that like charges repel each other.

### Step 2: Work Done by Forces
The work done by the force \( F_A \) you exert on \( q \) is denoted as \( W_A \), and the work done by the electrostatic force \( F_E \) is denoted as \( W_E \). Since you are moving \( q \) at a constant speed, the net work done on the charge must be zero. This means:

\[ W_A + W_E = 0 \quad \Rightarrow \quad W_A = -W_E \]

**Hint:** Think about the relationship between work done and the direction of forces.

### Step 3: Kinetic Energy Consideration
Since the charge \( q \) is moving at a constant speed, its kinetic energy does not change. Therefore, the change in kinetic energy \( \Delta KE = 0 \). According to the work-energy theorem:

\[ W_A + W_E = \Delta KE = 0 \]

This confirms that \( W_A = -W_E \).

**Hint:** Recall the work-energy theorem which relates work done to changes in kinetic energy.

### Step 4: Potential Energy Change
The potential energy \( U \) of the system is related to the work done by the electrostatic force. Since \( W_E \) is negative (as it is opposing the movement), the potential energy \( U \) of the system must decrease as \( q \) is moved away from the shell. Thus, we can say:

\[ \Delta U < 0 \quad \Rightarrow \quad U \text{ decreases} \]

**Hint:** Consider how potential energy changes with respect to the work done by conservative forces.

### Conclusion
From the analysis, we conclude that as the negative charge \( q \) is moved away from the negatively charged shell, the potential energy \( U \) of the system decreases. Therefore, the correct answer is:

**Option 4: \( U \) decreases.**