The plates of a parallel-plate capacitor are separated by a distance 2.02 x 10-3

Question

The plates of a parallel-plate capacitor are separated by a distance 2.02 x 10-3 m and filled with a slab of quartz of dielectric constant = 3.80. The volume of the capacitor is 6.06 x 10-5 m 3 and the plates have a surface charge density of magnitude = 2.00 x 10-7 C/m2 . a. How much charge stored in the capacitor? _________________________ b. What is the capacitance of this capacitor? _________________________ c. What is the potential difference between the plates of this capacitor? _________________________ d. What is the electric field strength between the capacitor plates? _________________________ e. What is the energy density stored in the capacitor?

Explanation / Answer

Distance between plates d = 2.02*10-3 m
Dielectric of slab = 3.8
Area of the plates = Volume /distance = 6.06*10-5 / 2.02*10-3 = 3*10-2 m2
Surface charge density = 2*10-7 C/m2
(a) Charge on the capacitor = Surface charge density*area
= 2*10-7 *3*10-2 = 6*10-9 C
(b) Capacitance C = KeoA/d = 3.8*8.85*10-12*3*10-2 /(2.02*10-3) = 50*10-11 F
(c) We know that
Q = CV
V = Q/C = 6*10-9 / 50*10-11 = 12 V
(d) E = Q/2Aeo = 2*10-7 /(2*8.85*10-12) = 0.11299*105 N/C = 11299.435 N/C
(e) Energy = (1/2)QV = 0.5*6*10-9*12 = 36*10-9 J

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