A parallel plate capacitor connected to a 2-V battery consists of two 1-cm square plates separated by 1 mm of air.How much work is done in inserting a sheet with a dielectric constant of 1.5 between the plates (in units of 10^-13 J)?

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by

C = (εA)/d

where ε is the permittivity of the medium between the plates (in this case, air), A is the area of each plate, and d is the separation between the plates.

Initially, with only air between the plates, the capacitance is

C = (ε_air A)/d = (8.85 x 10^-12 F/m * 0.01^2 m^2)/0.001m = 8.85 x 10^-14 F

When the sheet with the dielectric constant of 1.5 is inserted between the plates, the new capacitance becomes

C' = (ε_r ε_air A)/d = (1.5 * 8.85 x 10^-12 F/m * 0.01^2 m^2)/0.001m = 1.33 x 10^-13 F

The difference in energy stored in the capacitor before and after the dielectric is inserted is given by

ΔU = (1/2)C'V^2 - (1/2)CV^2

where V is the voltage across the plates. Plugging in the values, we get

ΔU = (1/2)(1.33 x 10^-13 F)(2 V)^2 - (1/2)(8.85 x 10^-14 F)(2 V)^2 = 8.85 x 10^-14 J

Multiplying by 10^13 to express in the desired units, we get

ΔU = 8.85 x 10^-1 J ≈ 8.85 x 10^-1 x 10^13 x 10^-13 = 8.85 x 10^0 x 10^-13 = 8.85 x 10^-13 J

Therefore, the answer is C) 8.85.

A Parallel Plate Capacitor Connected to a 2-V Battery Consists