Question

The electric field between the plates of a paper-separated (K=3.75) capacitor is 8.28

Answer 1

The seperation between the plates, d = 2.15 mm
= 2.15x10-3 m
The dielectric constant, k = 3.75
The electric field, E = 8.23x104 V/m
The charge on each plate, Q = 0.795 ?C
= 0.795x10-6 C
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SOLUTION:
The electric field between the plates is given by
E = V/d
Therefore, the potential difference across the plates is given by
V = Ed
= (8.23x104 V/m)(2.15x10-3 m)
= 176.94 V
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a)
The capacitance of the capacitor can be calculated as
C = Q/V
= (0.795x10-6 C)/(176.94 V)
= 4.49x10-9 F
= 4.49 nF
= 4.50 nF (approximately)
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b)
The capacitance for the parallel plate capcitor with dielctric is given by
C = k?0A/d
Therefore, the area of the plates is given by
A = Cd/k?0
= (4.49x10-9 F)(2.15x10-3 m)/(3.75)(8.85x10-12 C2/N.m2)
= 0.2910 m2
= 0.29 m2
= 0.30 m2 (approximately)