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 F
b) 7.775 * 10^-9
c) 7.444 * 10^-9 C
Explanation:
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 C
note:
calculation maybe wrong but method is correct. thanks