Problem: A Parallel Plate Capacitor A parallel plate capacitor is constructed of

Question

Problem: A Parallel Plate Capacitor A parallel plate capacitor is constructed of two square metal plates of side length l separated by a distance d. The capacitor is connected to a voltage source such that the potential difference between the plates is held constant at V .

a) If there is vacuum between the plates, determine the charge on the positive plate of the capacitor in terms of the given constants and 0. In terms of the same quantities, determine the energy stored in the capacitor.

b) A square sheet of an insulating solid of dielectric constant s, thickness d, and length greater than l is inserted between the plates of the capacitor such that it fills half the volume. In this situation, what is the total charge on the positive plate of the capacitor and what is the energy stored? Do these quantities increase or decrease compared to your answers for part (a)? Again use the constants given in your expressions. Hint: If you consider the capacitor to be broken into two capacitors (one with the dielectric and one without) how are they connected with respect to one another: in series or in parallel?

please show equations used and calculations thank you!!!

Explanation / Answer

Given
Area of the plate capacitor , A = l^2
plate seperation= d
Potential difference = V
a) Fzor a parallal plate capacitor
C = A*epsilon/d [ epsilon is permittivity of free space]
and Q = CV
so, Q = A*epsilon*V/d
Energy in capacitor = 0.5CV^2 = 0.5*A*epsilon*V^2/d
b) when half of the capacitor is filled with dielectric
we can assume this to be two capacitors in parallel, with half the cross section area
Capacitance of vaccum filled half = C/2
Capacitance of dielectric filled half = kC/2 [ where k is dielectric constant]
nety capacitance = C/2 + kC/2 = C(1 + k)/2
Q = C(1 + k)V/2 = A*epsilon(1+k)V/2d
E = 0.5C(1 + k)V^2/2 = 0.5*A*epsilon*(1+k)V^2/2d
we can see that Q increases ( as 1 + k > 2)
and E increases too

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