Parallel Plate Capacitor Problem, please provide detailed solutions for each par
ID: 1385424 • Letter: P
Question
Parallel Plate Capacitor Problem, please provide detailed solutions for each part and I will rate the best answer!
A parallel plates capacitor is made by placing two plates of area 5.90cm^2 a distance of 1.61 mm apart. a) Calculate the capacitance in pF assuming that there is air in between the plates. b) What would be the charge in pC, on each of the plates when we apply a voltage of 19.0 -V to the capacitor. c) How much energy, in Joules, is stored then in the capacitor? d) What if we fill the space between the plates with Bakelite whose dielectric constant K is 4.6, what would be the new capacitance in pF? Use 8.85 x 10^-12 F/m for e0 to the permittivity of free space or airExplanation / Answer
a)
capacitance , C = A*epsilon/d
C = 8.854 *10^-12 * 5.90 *10^-4/.00161
C = 3.24 *10^-12 F = 3.24 pF
capacitance is 3.24 pF
b)
as Q = CV
Q = 3.24 * 19
Q = 61.65 pC
c)
energy stored = 0.5 CV^2
energy stored = 0.5 * 3.24 *10^-12 * 19^2
energy stored =5.86 *10^-10 J
d)
New capacitance = k*c
New capacitance = 3.24 * 4.6
New capacitance = 14.9 pF
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