An electric charge `10^-3muC` is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at `(sqrt2, sqrt2)` and (2, 0) respectively. The potential difference between the points A and B will be

Options

A. (a) 4.5 volts
B. (b) 9 volts
C. (c) zero
D. (d) 2 volt

Explanation

To find the potential difference between points A and B due to a charge \( q = 10^{-3} \, \mu C \) placed at the origin (0, 0), we can follow these steps:

### Step 1: Understand the positions of points A and B
- Point A is located at \( ( \sqrt{2}, \sqrt{2} ) \).
- Point B is located at \( (2, 0) \).

### Step 2: Calculate the distances from the charge to points A and B
- The distance from the origin to point A, \( r_A \), can be calculated using the distance formula:
\[ r_A = \sqrt{(\sqrt{2} - 0)^2 + (\sqrt{2} - 0)^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \]

- The distance from the origin to point B, \( r_B \), is simply:
\[ r_B = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{4} = 2 \]

### Step 3: Calculate the electric potential at points A and B
The electric potential \( V \) due to a point charge is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r} \]
where \( q \) is the charge and \( r \) is the distance from the charge.

- For point A:
\[ V_A = \frac{1}{4 \pi \epsilon_0} \cdot \frac{10^{-3} \times 10^{-6}}{2} \]

- For point B:
\[ V_B = \frac{1}{4 \pi \epsilon_0} \cdot \frac{10^{-3} \times 10^{-6}}{2} \]

### Step 4: Calculate the potential difference \( \Delta V \)
The potential difference \( \Delta V \) between points A and B is given by:
\[ \Delta V = V_B - V_A \]
Since both potentials \( V_A \) and \( V_B \) are equal (as they are both calculated at the same distance from the charge), we have:
\[ \Delta V = 0 \]

### Final Answer
The potential difference between points A and B is \( 0 \).

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