Two uniformly charged spherical conductors A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be :

Solution

Physics sir · 3 steps

The wire makes both spheres one conductor, so set their potentials equal and the field goes as 1/r.

1. Equal potentials. Charge flows through the wire until both spheres are at the same potential. Treat each sphere as isolated, so V = kQ/r for each one.
2. Field from potential. At the surface, E = kQ/r² = 1/r·kQ/r = V/r. With V the same for both spheres, the field depends only on the radius.

   E_A/E_B = r_B/r_A = (10 mm)/(5 mm) = 2/1

3. So. The smaller sphere A has the stronger surface field, and E_A : E_B = 2 : 1, option (b).

The idea to keep. Connected conductors share one potential, so Q ∝ r, surface charge density σ ∝ 1/r and surface field E ∝ 1/r. That is why sharp points on a conductor have the strongest field.

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