Question

A parallel-plate capacitor of area 4 x 5 cm^2 is filled with mica (\episolon_r = 6). The
distance between the plates is 1 cm, and the capacitor is connected to a 100-V battery.
Calculate:
a) The capacitance of this capacitor
b) The free charge on the plates
c) The surface charge density due to the polarization charges
d) The field inside the mica. (what would the field be if the mica sheet were
withdrawn?)

Answer 1

here,

area = 4 * 5 cm^2 = 20 * 10^-4 m^2

k = 6

sepration between the plates , d = 1 cm = 0.01 m

potential difference , V = 100 V

a)

the capacitance of capacitor , C = K * e0 * area /d

C = 6 * 8.85 * 10^-12 * 20 * 10^-4 /0.01 F

C = 1.062 * 10^-11 F

b)

the free charge on the plates , Q = C * V

Q = 1.062 * 10^-11 * 100 = 1.062 * 10^-9 C

c)

The surface charge density due to the polarization charges , sigma = Q /area

sigma = 1.062 * 10^-9 /( 20 * 10^-4) = 5.31 * 10^-7 C/m^2

d)

the feild inside the mica , E = V /d

E = 100 /0.01 = 1 * 10^4 N/C