A rectangular parallel plate capacitor has 4 cm x 5 cm dimensions and the plates

Question

A rectangular parallel plate capacitor has 4 cm x 5 cm dimensions and the plates have a separation of 2 cm. The capacitor is connected to a source that supplies 100 v calculate:

(a) The maximum charge in the capacitor

(b) The energy stored in the capacitor

(c) The electric field in the capacitor

(d) The density of electric power of the capacitor

(e) The capacitance of the capacitor if between the parallel plates is introduced a dielectric with a dielectric constant of 68.

(f) If a dielectric is introduced between the plates of the capacitor and has a dielectric constant of 100, the charge induced on the dielectric.

*Show all the steps and diagram

Explanation / Answer

Given

Capacitor of rectangular parallel plate of dimensions 4cm X 5 cm, Area A = 20 cm^2 = 0.2 m^2

separation is d = 2 cm = 0.02 m

power supply of V = 100 V

a) maximum charge is Q = C*V , but C is capacitance of the capacitor is

C = epsilon not*A/d = 8.854*10^-12*0.2/0.02 F = 8.854*10^-11 F = 88.54 pF

now Q = 88.54*10^-12*100 = 8.854 *10^-9 C = 8.854 nC

b)energy stored in the capcitor is U = 0.5*C*V^2

U = 0.5*88.54*10^-12*100^2 J

U = 4.427*10^-7 J

c)electric field in the capacitor is E = sigma /epsilon not

E = (Q/A)/epsilon not

E = (8.854*10^-9/0.2)/(8.854*10^-12) N/C

E = 5000 N/C

d) energy density of the electric power is u = 0.5*epsilon not*E^2

u = 0.5*(8.854*10^-12)(5000)^2 FV^2/m^3

e) if the dielectric material placed in the capacitor then

C' = k*C = 68*88.54*10^-9 F = 6.02072*10^-6 F

f)

if k = 100 placed between plates of capacitor then the charge is

Q = C*V

Q' = k*c*V

Q' = 100*88.54*10^-9*100C= 0.0008854 C 8.854*10^-4 C

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