# Capacitance Dielectric Problem

## Homework Statement

A parallel plate capacitor is constructed by placing square metal sheets of side length 30 cm on either side of a sheet of polythene which is 2mm thick
If the permeability of the polythene is 2.2 *10^-11 F m^-1, what is the capacitance of the system
(ii)If a 50V power supply is connected across the plates, calculate the charge on the plates and the energy stored in the capacitor
(iii) If the polythene is withdrawn from the capacitor until only half of the space bewtween the plates is occupied by it, what is the charge on the plates and the energy stored in this case?

C=Q/V
C= epsi*A/D
E= 1/2*C*V^2

## The Attempt at a Solution

I think i could do part 1 and 2 alright just using the standard equations
(i)C = (2.2*10^-11)*(9.o*10^-2)/(2.0*10^-3) = 9.9*10^-10

(ii)Q= CV
Q= (9.9*10^-10)*50 = 4.95*10^-8
E = 0.5(9.9*10^-10)*(50^2)= 1.24*10^-6

(iii)I dont know what to do for this part, Will the voltage decrease and the charge also?
any tips or hints would be very much appreciated
thanks debs

## Answers and Replies

Päällikkö
Homework Helper
Well what withdrawing the polythene does is that you can think of the problem as if you had two parallel capacitors, one with polythene, one without. Use the second equation (C = epsi*A/D) to work out the details. Voltage remains constant as it's supplied from a battery.

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(simultaneous post)

I'm thinking that the way to solve the latter problem is to treat them as capacitors in series, not parallel, where they add like resistors in parallel.

1/Ceq=d1/Ake+d2/Ae where e =permitivity constant in vacuum d1 and d2 are respective thicknesses and K the dielectric value for the plastic film.

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marcusl
Gold Member
Why would they be in series? The same plates serve for both capacitors.

Why would they be in series? The same plates serve for both capacitors.

right but the electric field crosses two distinctly different regions, and the addition of these leads to what I suggested iirc. Might be wrong, however. Its been a long time since I took physics....I should think derivation using gauss' law would not be too difficult in any event. Maybe when I come back I will poke around at this problem some more if no one has commented definitively.

marcusl
Gold Member
right but the electric field crosses two distinctly different regions, and the addition of these leads to what I suggested iirc. Might be wrong, however.
Yeah, that would only be correct if the spacing between plates was greater than 2mm, so the dielectric consisted of both plastic and air.

i thought that was exactly the case, maybe I misread the problem. BTW where does the limit 2mm come from--is this an empirical constant?

marcusl