Question

The plates of a spherical capacitor have radii 49.2 mm and 51.4 mm. (a) Calculate the capacitance. (b) What must be the plate area of a parallel-plate capacitor with the same plate separation and capacitance?

Answer 1

Given

Spheirical capacitor with radii

let r1 inner = 49.2 mm, outer r2 = 51.4 mm  

a)

we know that the relation between the capacitance ,charge and potentila is

Q = C*V ==> C = Q/V = Q/(V1- V2)

where V1,V2 are the potential of the the surfaces

V1 = k*q/r1 , v2 = kq/r2 and k = 1/(4pi*)

also

the capacitance of a spherical capacitor with radii r1,r2 is  

C = (4pi)(r1*r2)/((r2-r1)

substituting the values

C = (4pi*8.854*10^-12*49.2*10^-3*51.4*10^-3)/((51.4*10^-3)-(49.2*10^-3)) F

C = 127.8954 pF

b)

we knwo that the cpacitance of a prallel plate capacitor with area A and separated by a distance d is  

C = *A/d

A = C*d/()

A = ((4pi)(r1*r2)/((r2-r1))(d)/()

A = (4pi*r1*r2)/((r2-r1)(d))

Area of the parallelplate capacitor is A = (4pi*r1*r2)/((r2-r1)(d)) = 14.44493*d m^2

Area of the parallelplate capacitor is A = (4pi*r1*r2)/((r2-r1)(d)) = 14.44493*d m^2