Question

A parallel plate capacitor has circular plates of diameter 5 cm and separation 2 mm. The space between the plates is filled with a material of dielectric constant K = 3. The charges on the plates are ± q. The charge is given by q = 27 × 10−9 C.

Find: (i) the charge; (ii) the capacitance; (iii) the potential difference between the plates; (iv) the magnitude of the electric field between the plates; (v) the electric energy density between the plates; and (vi) the total electric energy

Answer 1

The charge is given to us as q=27*10^-9C

B):the capacitance of circular capacitor is given by

C=epsilon° x area /separation

Epsilon°= dielectric constant x epsilon= k x epsilon°.

Area(m²) =3.14*(2.5*10-²)²=3.14*6.25*10-⁴=19.625*10-⁴

Separtion, l=2mm=2*10-³m

So capacitance=k x epsilon° x area/separation

Capacitance=3*8.85*10^-12*19.625*10-⁴/2*10-³

Capacitance=260.52*10^-13 farad.=2.60*10^-11 farad.

C) :as charge=capacitance x voltage

Q=cv

27*10^-9=2.60*10^-11*V

V=27*10^-9/2.60*10^-11

V=10.385*10² volts.

D):as E= - dv/dr

Where E=electric field, dv=potential difference and dr=distance

E=-dv/dr= - 10.385*10²/2*10-³

E=5.192*10^5 c/m.

E) :energy density in capacitor is given by U

U=1/2 epsilon°xE²

U=1/2*3*8.85*10^-12*(5.192*10^5)²

U=0.5*8.85*10^-12*26.958*10^10

U=119.3*10-²

F):total electric energy is given by

Energy =1/2CV²

Energy =0.5*2.6*10^-11*(10.385*10²)²

Energy =0.5*2.6*10^-11*107.85*10⁴

Energy=140.202*10-^6

Energy =1.40*10-⁴ joule.