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 determine the potential difference between points P and Q caused by a charge at the origin, the formula for the electric potential of a point charge is applied:

\(V = \frac{kQ}{r}\)

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

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

The potential difference is 0 V. This is because the potentials at two points equidistant from the same charge are equal, resulting in a zero potential difference.