The space between two concentric conducting spherical shells of radii b = 1.70 cm and a = 1.20 cm is filled with a substance of dielectric constant ? = 27.0. A potential difference V = 64.5 V is applied across the inner and outer shells.

(a) Determine the capacitance of the device.

nF

(b) Determine the free charge q on the inner shell.

nC

(c) Determine the charge q’ induced along the surface of the inner shell.

nC

Answer:a) C = 1.065 * 10^-10 Fb) 7.775 * 10^-9c)7.444 * 10^-9 CExplanation:A spherical capacitor, with inner radius of a = 1.2 cm and outer radius

of b = 1.7 cm is filled with a dielectric material with dielectric constant of

K = 27 and connected to a potential difference of V = 64.5 V.

(a)The capacitance of a filled air spherical capacitor is given by equation :C = 4*π*∈o*(a*b/b-a)

if the capacitor is filled with a material with dielectric constant K, we need

to modify the capacitance as ∈o —->k∈o , thus:

C = 4*π*∈o*(a*b/b-a)

substitute with the given values to get:

C = 4*π*(27)*(8.84*10^-12)[(1.2*10^-2)*(1.7*10^-2)/(1.7*10^-2)-(1.2*10^-2)*]

C = 1.065 * 10^-10 F(b)The charge on the capacitor is given by q = CV, substitute to get:q = (1.065 * 10^-10)*64.5 V

= 7.775 * 10^-9(c)The induced charge on the dielectric material is given by equation as:q’ = q(1-1/k)

substitute with the given values to get:

q’ = (7.775 * 10^-9)*(1-1/27)

= 7.444 * 10^-9 Cnote:calculation maybe wrong but method is correct. thanks