A metallic sphere has a charge of `10 mu C`. A unit negative charge is brought from `A` to `B` both `100 cm` away from the sphere but `A` being east of it while `B` being on west. The net work done is

Options

A. Zero
B. `2//10` joule
C. `-2//10`joule
D. `-1//10` joule

Explanation

To solve the problem step by step, we will analyze the situation involving the metallic sphere and the movement of the unit negative charge from point A to point B.

### Step 1: Understand the Setup
We have a metallic sphere with a charge of \( Q = 10 \, \mu C = 10 \times 10^{-6} \, C \). A unit negative charge \( q = -1 \, C \) is moved from point A to point B. Both points A and B are located at a distance of \( r = 100 \, cm = 1 \, m \) from the center of the sphere, with point A being east and point B being west of the sphere.

**Hint:** Identify the positions of the charges and their distances from the sphere.

### Step 2: Calculate the Electric Potential at Points A and B
The electric potential \( V \) due to a point charge (or a charged sphere) at a distance \( r \) is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, N \cdot m^2/C^2 \)).

For both points A and B, the distance \( r \) is the same (1 m), so we can calculate the potential at both points:
\[ V_A = \frac{kQ}{1} = kQ \]
\[ V_B = \frac{kQ}{1} = kQ \]

**Hint:** Remember that the electric potential is the same at equal distances from a charged sphere.

### Step 3: Determine the Work Done
The work done \( W \) when moving a charge in an electric field is given by:
\[ W = q(V_B - V_A) \]
Since we have already established that \( V_A = V_B \), we can substitute:
\[ W = -1 \, C \cdot (V_B - V_A) = -1 \, C \cdot (0) = 0 \]

**Hint:** The work done is dependent on the change in potential energy, which is zero if the potentials are equal.

### Step 4: Conclusion
The net work done in moving the unit negative charge from point A to point B is:
\[ W = 0 \]
Thus, the answer to the question is that the net work done is zero.

**Final Answer:** The net work done is \( 0 \, J \).