A parallel plates capacitor is made by placing two plates of area 5.40cm 2 a dis

ID: 1386166 • Letter: A

Question

A parallel plates capacitor is made by placing two plates of area 5.40cm2 a distance of 1.76mm 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 23.0-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 ? is 7.3, what would be the new capacitance in pF?

Use 8.85 x 10-12 F/m for ?o, the permittivity of free space or air.

A) C = eo*A/d

A is the area of the plates = 5.4*10^-4 m^2

and d is the distance between the plates = 1.76*10^-3 m

Capacitance C = eo*A/d = (8.85*10^-12*5.4*10^-4)/(1.76*10^-3) = 2.71 pF

B) C = Q/V
Cahrge Q = C*V = 2.71*10^-12*23 = 62.33 pC

C) Energy stored is U = 0.5*Q*V = 0.5*62.33*10^-12*23 = 7.167*10^-10 J

D) Cnew = k*Cold = 7.3*2.71*10^-12 = 19.78 pF

k is dielectric constant

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