1. Calculate the capacitance in farads (F) of two parallel plates, each with an

Question

1. Calculate the capacitance in farads (F) of two parallel plates, each with an area of 2.0 m2 and separated by a gap of 1.0 mm (1 x 103 m). Assume that there is air in the gap between the plates. How much charge is stored on the positive plate if the electric potential between the plates is 12 V? If the 12 V battery which charges the capacitor remains connected to the capacitor and a piece of paper is slipped in the gap between the plates, will the amount of positive charge on the positive plate increase, decrease or remain the same? Why? What happens to the capacitance? What happens to the voltage? (Need work)

Explanation / Answer

Capacitance C is given by

C= K*Eo*A/D, where Eo= 8.854x10-12

where:

K is the dielectric constant of the material,
A is the overlapping surface area of the plates,
d is the distance between the plates, and
C is capacitance

Here, K of air=1.00054

C=8.854x10-12x1.00054x2/10^-3

C=1.77x10^-8 F

So, Capacitance is 1.77x10^-8 F

Charge stored on positive plate=W

W=CV^2/2

W=1.77x10^-8x12x12/2

W=12.7x10^-7J

If paper is inserted( a dielectric is inserted,) then C increases and thus the energy decreases because someone has to insert or remove the dielectric and that will require work (either positive or negative depending on the situation). Work is where the energy goes.

Capacitance increases and Voltage decreases

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