Problem 17.55 A homemade capacitor is assembled by placing two 10-in. pie pans 5

Question

Problem 17.55 A homemade capacitor is assembled by placing two 10-in. pie pans 5 cm apart and connecting them to the opposite terminals of a 6-V battery. Part A Estimate the capacitance. Express your answer using one significant figure. C = F SubmitMy AnswersGive Up Part B Estimate the charge on each plate. Express your answer using one significant figure. Q = C SubmitMy AnswersGive Up Part C Estimate the electric field halfway between the plates. Express your answer using one significant figure. E = V/m SubmitMy AnswersGive Up Part D Estimate the work done by the battery to charge the plates. Express your answer using one significant figure. W = J SubmitMy AnswersGive Up Part E Which of the above values change if a dielectric is inserted? Which of the above values change if a dielectric is inserted? Capacitance. Charge. Electric field. Work done by the battery.

Explanation / Answer

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

radius = 5 inch = 0.127 m

A = pi x r^2 = .056m^2

C = (8.9e-12)(0.056) / (0.05) = 9.968e-12 = 9.97 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 =
= 9.97 F x6 V x (1 coulomb / (1 F x 1 V))
= 59.82p C

c) Field = V/d = 6/0.05 = 120 v/m
Electric field is V/d
assume constant e field

d)
E = ½CV² = ½(9.97p)(6)² = 179.46 pJ

e) if the battery is still connected...
charge and work and capacitance changes
If the battery is not connected
voltage and capacitance changes, and field

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