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 solve the problem, we need to find the potential difference between the points (0, a) and (a, 0) due to a charge of 10 μC placed at the origin (0, 0) of the x-y coordinate system.

### Step-by-Step Solution:

1. **Identify the Charge and its Position**:
- We have a point charge \( Q = 10 \, \mu C = 10 \times 10^{-6} \, C \) located at the origin (0, 0).

2. **Determine the Points**:
- The two points we are considering are:
- Point A: (0, a)
- Point B: (a, 0)

3. **Calculate the Distance from the Charge to Each Point**:
- The distance from the charge at the origin to point A (0, a) is:
\[ r_A = \sqrt{(0 - 0)^2 + (a - 0)^2} = \sqrt{a^2} = a \]
- The distance from the charge at the origin to point B (a, 0) is:
\[ r_B = \sqrt{(a - 0)^2 + (0 - 0)^2} = \sqrt{a^2} = a \]

4. **Use the Formula for Electric Potential**:
- The electric potential \( V \) due to a point charge is given by:
\[ V = \frac{kQ}{r} \]
- Where \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, N m^2/C^2 \)).

5. **Calculate the Potential at Each Point**:
- For point A:
\[ V_A = \frac{kQ}{r_A} = \frac{k \cdot 10 \times 10^{-6}}{a} \]
- For point B:
\[ V_B = \frac{kQ}{r_B} = \frac{k \cdot 10 \times 10^{-6}}{a} \]

6. **Find the Potential Difference**:
- The potential difference \( V_{AB} \) between points A and B is given by:
\[ V_{AB} = V_A - V_B \]
- Since \( V_A \) and \( V_B \) are equal:
\[ V_{AB} = \frac{k \cdot 10 \times 10^{-6}}{a} - \frac{k \cdot 10 \times 10^{-6}}{a} = 0 \]

### Final Answer:
The potential difference between the two points (0, a) and (a, 0) is **0 volts**.