Question

Two parallel conducting plates have equal and opposite charges. Consider the area of conducting place is 2.5 cm^2 and the capacitor is filled with 1.8 m thick dielectric material which has  K=3.60  dielectric constant. The resultant electric field in the dielectric is 1.20×10⁶ V/m.

a-Find the magnitude of the charge density σ on the conducting plate.

b-Calculate the magnitude of the charge density σ1 on the surfaces of the dielectric.

c-Determine the total electric-field energy U stored in the capacitor.

Answer 1

a)Resultant electric field E=/(k)

=surface charge density on plates

k=dielectric constant=3.6

=8.85X

E=1.2XV/m

E=/(k)

=Ek=1.2XX3.6X8.85X=3.82XC/m2

b)Charge density on dielectric 1=(1-1/k)=3.82XX(1-1/3.6)=2.76XC/m2

c)Total enrgy in capacitor E=C/2

C=capacitance=Ak/d

A=area of plate=2.5cm2=2.5Xm2

d=distance between plates=1.8m (I think it is 1.8m , but d is very small in capacitor in unit of mm)

C=capacitance=Ak/d=2.5XX8.85XX3.6/1.8=44.25XF

V=voltage=Ed=1.2XX1.8=2.16Xvolt

Total enrgy in capacitor E=C/2=44.25XX/2=1.03XJ