Electric Potential — Practice Quiz

4.5 × 10^6 N/C, 4.5 × 10^6 V, 1.5 × 10^7 V

Radius is 0.30 m. Outside the sphere, it behaves like a point charge at the centre. At r = 1.0 m, E = kq/r^2 = 9 × 10^9 × 500 × 10^-6 / 1^2 = 4.5 × 10^6 N/C and V = kq/r = 4.5 × 10^6 V. Inside a hollow conducting sphere, electric field is zero and potential is uniform, equal to the surface potential. Surface potential is V = kq/R = 9 × 10^9 × 500 × 10^-6 / 0.30 = 1.5 × 10^7 V.

5.4 mm and 795 V

For one drop, V = kq/r, so r = kq/V = 9 × 10^9 × 3.0 × 10^-10 / 500 = 5.4 × 10^-3 m = 5.4 mm. When two identical drops combine, the charge doubles and radius becomes 2^(1/3) times the original. So new potential is V' = 2^(2/3)V = 1.59 × 500 = 795 V.

1.1 × 10^5 V and 1.1 × 10^17 V/m

Charge on the nucleus is q = 79e = 79 × 1.6 × 10^-19 C. Potential is V = kq/r = 9 × 10^9 × 79 × 1.6 × 10^-19 / 10^-12 = 1.1 × 10^5 V. Electric field is E = kq/r^2 = V/r = 1.1 × 10^5 / 10^-12 = 1.1 × 10^17 V/m.

8.85 mm

For a sheet of charge, E = σ/(2ε0). Here E = 10^-8 /(2 × 8.85 × 10^-12) = 564 N/C approximately. Using E = ΔV/Δr, we get Δr = ΔV/E = 5/564 ≈ 8.85 × 10^-3 m = 8.85 mm.