An electric charge 10⁻⁶ μ C is placed at origin (0,0) m of X-Y co-ordinate system. Two points P and Q are situated at (√3, √3) m and (√6, 0) m respectively. The potential difference between the points P and Q will be :

Solution

Physics sir · 3 steps

Check the distances from the charge first: P and Q are both √6 m away, so nothing changes.

1. Point charge potential. A point charge q produces potential V = kq/r, where r is the distance from the charge. Only the distance matters, not the direction.
2. Distances from origin. For P, r_P = √((√3)² + (√3)²) = √6 m. For Q, r_Q = √((√6)² + 0²) = √6 m.

   V_P - V_Q = kq(1/r_P - 1/r_Q) = kq(1/√6 - 1/√6) = 0

3. So. P and Q lie on the same equipotential circle of radius √6 m around the charge. The potential difference is 0 V, option (c). The size of the charge never enters.

The idea to keep. Around a single point charge, the equipotentials are spheres centred on it. Any two points at equal distance from the charge have zero potential difference, whatever the charge is.

Answer from the NTA final answer key. Solution by PrepWiser.