A charge of 10`mu`C is placed at the origin of x-y coordinate system. The potnetial difference between two point (0, a) and (a, 0) in volt will be

Options

A. `(9 xx 10^(-4))/(a ) `
B. `(9 xx 10^(4))/(8 sqrt(2))`
C. `(9x 10^(4))/(2a) `
D. zero

Answer

To find the potential difference between the two points (0, a) and (a, 0) due to a charge of 10 µC placed at the origin, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Points and Charge**:
- We have a charge \( Q = 10 \, \mu C = 10 \times 10^{-6} \, C \) located at the origin (0, 0).
- We need to find the potential difference between the points \( P(0, a) \) and \( Q(a, 0) \).

2. **Calculate the Distance from Charge to Points**:
- The distance from the charge at the origin to point \( P(0, a) \) is:
\[ r_P = \sqrt{(0 - 0)^2 + (a - 0)^2} = a \]
- The distance from the charge at the origin to point \( Q(a, 0) \) is:
\[ r_Q = \sqrt{(a - 0)^2 + (0 - 0)^2} = a \]

3. **Calculate the Electric Potential at Each Point**:
- The electric potential \( V \) due to a point charge is given by the formula:
\[ V = \frac{kQ}{r} \]
- For point \( P(0, a) \):
\[ V_P = \frac{k \cdot 10 \times 10^{-6}}{a} \]
- For point \( Q(a, 0) \):
\[ V_Q = \frac{k \cdot 10 \times 10^{-6}}{a} \]

4. **Calculate the Potential Difference**:
- The potential difference \( V_{P} - V_{Q} \) is:
\[ V_{P} - V_{Q} = \frac{k \cdot 10 \times 10^{-6}}{a} - \frac{k \cdot 10 \times 10^{-6}}{a} \]
- This simplifies to:
\[ V_{P} - V_{Q} = 0 \]

5. **Conclusion**:
- The potential difference between the two points \( (0, a) \) and \( (a, 0) \) is \( 0 \, V \).

### Final Answer:
The potential difference between the points \( (0, a) \) and \( (a, 0) \) is \( 0 \, V \).

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