The electric potential `V` in volts in a region of space is given by `V = ax^(2) + ay^(2) + 2azy`. The work done by field when a `2C` test charge moves from point `(0, 0, 0.1m)` to origin is `5 xx 10^(-5) J` . Determine `a`.

Options

A. `1000`
B. `650`
C. `1500`
D. `1250`

Explanation

To solve the problem, we need to determine the value of \( a \) given the electric potential \( V \) and the work done by the electric field when a test charge moves from a specific point to the origin.

### Step-by-Step Solution:

1. **Identify the Electric Potential Expression**:
The electric potential \( V \) is given by:
\[ V = ax^2 + ay^2 + 2azy \]

2. **Determine the Initial and Final Positions**:
The test charge moves from the point \( (0, 0, 0.1 \, \text{m}) \) to the origin \( (0, 0, 0) \).

3. **Calculate the Potential at the Initial Position**:
At the initial position \( (0, 0, 0.1) \):
- \( x = 0 \)
- \( y = 0 \)
- \( z = 0.1 \)

Substitute these values into the potential equation:
\[ V_{\text{initial}} = a(0)^2 + a(0)^2 + 2a(0)(0.1) = 0 + 0 + 0 = 0 \]

4. **Calculate the Potential at the Final Position (Origin)**:
At the final position \( (0, 0, 0) \):
- \( x = 0 \)
- \( y = 0 \)
- \( z = 0 \)

Substitute these values into the potential equation:
\[ V_{\text{final}} = a(0)^2 + a(0)^2 + 2a(0)(0) = 0 + 0 + 0 = 0 \]

5. **Calculate the Potential Difference**:
The potential difference \( \Delta V \) when moving from the initial position to the final position is:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 0 - 0 = 0 \]

6. **Use the Work Done Formula**:
The work done \( W \) by the electric field when moving a charge \( q \) through a potential difference \( \Delta V \) is given by:
\[ W = q \Delta V \]
Given that \( W = 5 \times 10^{-5} \, \text{J} \) and \( q = 2 \, \text{C} \):
\[ 5 \times 10^{-5} = 2 \cdot \Delta V \]

7. **Solve for the Potential Difference**:
Rearranging the equation gives:
\[ \Delta V = \frac{5 \times 10^{-5}}{2} = 2.5 \times 10^{-5} \, \text{V} \]

8. **Relate the Potential Difference to \( a \)**:
From the potential difference calculated earlier, we know:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 0 - (0.2a) = -0.2a \]
Setting this equal to the calculated \( \Delta V \):
\[ -0.2a = 2.5 \times 10^{-5} \]

9. **Solve for \( a \)**:
Rearranging gives:
\[ a = -\frac{2.5 \times 10^{-5}}{0.2} = -1.25 \times 10^{-4} \, \text{V/m}^2 \]

### Final Answer:
\[ a = 1.25 \times 10^{-4} \, \text{V/m}^2 \]