A conducting sphere of radius R is charged to a potential V . It is then connected by a long, thin conducting wire to an initially uncharged conducting sphere of radius 2R . After electrostatic equilibrium is reached, what is the electric field on the surface of the larger sphere?

Step-by-step solution

Let k = 1 4 ₀ . The initial charge on the first sphere is Q = V R k . When the two spheres are connected, they share charge until they reach a common potential V' . The common potential is given by: V' = Total Charge Sum of Capacitances = Q C₁ + C₂ Since the capacitance of a spherical conductor is proportional to its radius ( C = R k ): V' = V R k R k + 2R k = V R 3R = V 3 The electric field on the surface of a conducting sphere is related to its surface potential by E = V_ surface Radius . For the larger sphere (r