Question

An electric charge \(10^{-6} \, \mu C\) is placed at the origin (0, 0) of an X-Y coordinate system. Two points P and Q are situated at \((\sqrt{3}, \sqrt{3}) \, \text{mm}\) and \((\sqrt{6}, 0) \, \text{mm}\) respectively. The potential difference between the points P and Q will be:

• A. 0 V
• B. \( \sqrt{6} \) V
• C. \(\sqrt{3} \) V
• D. 3 V

Answer 1

The Correct Option is A

To find the potential difference between points P and Q due to a charge placed at the origin, we use the formula for electric potential due to a point charge:

\(V = \frac{kQ}{r}\) 

where \(V\) is the electric potential, \(k\approx8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\) is the Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge.

1. Calculate the distance from the charge at the origin to point P \((\sqrt{3}, \sqrt{3}) \, \text{mm}\): \(r_P = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{6} \, \text{mm} = \sqrt{6} \times 10^{-3} \, \text{m}\)
2. Determine the distance from the charge to point Q \((\sqrt{6}, 0) \, \text{mm}\): \(r_Q = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6} \, \text{mm} = \sqrt{6} \times 10^{-3} \, \text{m}\)
3. Since both points P and Q are equidistant from the origin, the potentials at P and Q are the same: \(V_P = \frac{k \cdot 10^{-6} \cdot 10^{-6}}{\sqrt{6} \times 10^{-3}} = V_Q\)
4. Thus, the potential difference between the two points P and Q is: \(\Delta V = V_P - V_Q = 0 \, \text{V}\)

The correct answer is 0 V. The potential at two equidistant points from the same charge is equal, leading to a potential difference of zero.

Answer 2

The potential difference between two points P and Q due to a point charge Q is given by:

\[ \Delta V = KQ \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

where

\[ r_1 = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{6} \quad \text{and} \quad r_2 = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6} \]

Since \(r_1 = r_2\), the potential difference is zero:

\[ \Delta V = 0 \]