In: Physics
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.
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
=1.2X
X3.6X8.85X
=3.82X
C/m2
b)Charge density on dielectric
1=
(1-1/k)=3.82X
X(1-1/3.6)=2.76X
C/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=2.5X
X8.85X
X3.6/1.8=44.25X
F
V=voltage=Ed=1.2XX1.8=2.16X
volt
Total enrgy in capacitor E=C/2=44.25X
X
/2=1.03X
J