(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d. (b) Two charged spherical conductors of radii R1 and R2 when connected by a conducting wire acquire charge q1 and q2 respectively. Find the ratio of their surface charge densities in terms of their radii.

(a)Electric field on the inside of the parallel plate capacitor due to one of the plates is given by: E1 = σ/2εo = Q/2Aεo
Total electric field due to the two plates is given by: E = 2E1 = Q/Aεo
Potential difference between the plates is given by: V = Ed
V = Qd/Aεo
By definition, C = Q/V = Aεo/d
(b)After connection, both spheres are at the same potential.
V1 = V2
Kq1/R1 = Kq2/R2
q1/q2 = R1/R2
Surface charge density for a spherical conductor is given by:
σ = Q/S
Hence,
σ1 = q1/4πR1²
σ2 = q2/4πR2²
σ1/σ2 = (q1/4πR1²) / (q2/4πR2²) = q1R2²/q2R1² = (R1/R2)(R2²/R1²) = R2/R1