1. A parallel plate capacitor has plates that are 1 m by 1m (w 1.0 m and 1-10 m

Question

1. A parallel plate capacitor has plates that are 1 m by 1m (w 1.0 m and 1-10 m in picture below) separated by a distance of 104 m (0.1 mm). Initially there is no dielectric between the plates. A 6volt battery is connected across the plates a) Determine the capacitance of this capacitor. b) Determine the Q and on the positive plate. Is this charge free or bound? c) Determine the electric field in between the plates. d) Determine the electric displacement between the plates. e) Determine the energy stored in this capacitor.

Explanation / Answer

part a:

capacitance=k*epsilon*area/distance between plates

=3*8.85*10^(-12)*1*1/10^(-4)

=2.655*10^(-7) F

part b:

total charge=capacitance*voltage

=2.655*10^(-7)*6=1.593*10^(-6) C

free charge==capacitance without dielectric*voltage

=(8.85*10^(-12)*1*1/10^(-4))*6

=5.31*10^(-7) C

bound charge=total charge-free charge=

=1.062*10^(-6) C

part c:

E in the dielectric=potential difference/distance between plates

=6 volts/(10^(-4) m)

=6*10^4 V/m

part d:

D=epsilon*E

=k*epsilon_0*E

=3*8.85*10^(-12)*6*10^4

=1.593*10^(-6) C/m^2

part e:

P=D-(epsilon_0*E)

=1.593*10^(-6)-(8.85*10^(-12)*6*10^4)

=1.062*10^(-6) C/m^2

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