Question

A homemade capacitor is assembled by placing two 9-in. pie pans 4cm apart and connecting them to the opposite terminals of a 6-Vbattery.

Part A:

Estimate the capacitance.

C= ?? V

Part B:

Estimate the charge on each plate.

Q= ?? C

Part C:

Estimate the electric field halfway between the plates.

E = V/m

Part D:

Estimate the work done by the battery to charge the plates.

W= ?? J

Answer 1

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A homemade capacitor is assembled by placing two 9-in. pie pans 5 cm apart and connecting them to the opposite terminals of a 9-V battery. Estimate (a) the capacitance, (b) the charge on each plate, (c) the electric field halfway between the plates, and (d) the work done by the battery to charge the plates. (e) Which of the above values change if a dielectric is inserted?

a)

use the equation

C = E * A/d

where C is capacitance in farads
E = permitivity of insulator
A = area of each plate in m^2
d is separation in meters

for air, E = Er x Eo = 1.00054 x 8.85419 x 10^-12 F/m
A = pi x r^2 = pi x 9 in^2 x (2.54 cm/in)^2 x (1 m / 100 cm)^2
= 1.82415 x 10^-2 m^2
d = 8 cm x 1 m /100 cm = 8 x 10^-2 m

so that C = 2 x 10^-12 F = 2 pF

b)

use C = Q/V

where C is capacitance from a)
Q is charge on each plate
V is potential

Q = C x V =
= 2 x 10^-12 F x 9 V x (1 coulomb / (1 F x 1 V))
= 18 x 10^-12 C = 1 pC

c)

Electric field is V/d
assume constant e field
so halfway between plates would be 1/2 x V/d
= 1/2 x 9 V x .08 m = .36 V/m

d)

dW = Q/C x dQ
integral (dW) = integral 0 to Q of (Q/C dQ)
W = 1/2 Q^2 / C

but C = Q/V so Q^2 = C^2/V^2

so
W = 1/2 C/V^2

W = work ( in joules)
C = capacitance ( in Farads)
V = volts

W = 1/2 x 2 x 10^-12 x 9 = 9 x 10^-12 Joules