A homemade capacitor is assembled by placing two 10 in pie pans 11 cm apart and

ID: 1404702 • Letter: A

Question

A homemade capacitor is assembled by placing two 10 in pie pans 11 cm apart and connecting them to the opposite terminals of a 9 V battery.

(a) Estimate the capacitance? (pF)

(b) Estimate the charge on each plate? (pC)

(d) Estimate the work done by the battery to charge the plates? (J)

(e) Which of the above values change if a dielectric is inserted? (Select all that apply.)

1. work done by the battery 2. electric field

3. capacitance 4. charge

given,

diameter = 10 inch or 25.4 cm

distance between the plates = 11 cm

voltage = 9 V

so,

area of the plate = pi * radius^2

area of the plate = pi * (0.254)^2

area of the plate = 0.2027 m^2

capacitance = k * area / distance

capacitance = 8.854 * 10^-12 * 0.2027 / 0.11

capacitance = 1.632 * 10^-11 F

charge = capavitance * voltage

charge = 1.632 * 10^-11 * 9

charge = 1.468 * 10^-10 C

electric field = voltage / distance

so at halfway distance will be 0.11 / 2 so

electric field = 9 / (0.11/2)

electric field = 163.64 N/C

work done = 0.5 * C * V^2

work done = 0.5 * 1.632 * 10^-11 * 9^2

work done = 6.61 * 10^-10 J

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