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. 9 V
B. zero
C. 2 V
D. 3.5 V

Answer

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

### Step 1: Identify the coordinates 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 potential \( V \) due to a point charge is given by the formula:
\[ V = \frac{k \cdot q}{r} \]
where:
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q \) is the charge,
- \( r \) is the distance from the charge to the point.

#### Distance from the charge to point A:
Using the distance formula:
\[ r_A = \sqrt{(\sqrt{2} - 0)^2 + (\sqrt{2} - 0)^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \]

#### Distance from the charge to point B:
Using the distance formula:
\[ r_B = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{4} = 2 \]

### Step 3: Calculate the potentials at points A and B
Now, we can calculate the potentials:
\[ V_A = \frac{k \cdot q}{r_A} = \frac{k \cdot (10^{-3} \times 10^{-6})}{2} \]
\[ V_B = \frac{k \cdot q}{r_B} = \frac{k \cdot (10^{-3} \times 10^{-6})}{2} \]

### Step 4: Calculate the potential difference \( \Delta V \)
The potential difference between points A and B is given by:
\[ \Delta V = V_B - V_A \]
Since both potentials are equal:
\[ \Delta V = \frac{k \cdot (10^{-3} \times 10^{-6})}{2} - \frac{k \cdot (10^{-3} \times 10^{-6})}{2} = 0 \]

### Conclusion
The potential difference between points A and B is:
\[ \Delta V = 0 \, \text{V} \]

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