# Gauss' Law Question

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#1
>A parallel plate capacitor with dielectric (as above), together with its dimensions. Its plates are square. The capacitance is given by the usual formula, Where:
A= 0.2 t= 0.1mm = 2.2 F/m.

> (a) Assuming the top plate is positively charged, make your own copies of the diagram (without the labeling) and on them sketch respectively, (i) Electric field lines (go from positive plate to negative) and (ii) the lines of electric flux density (go perpendicular to field lines and curve at the edges of each plate.

>(b) If the capacitor is charged to 500 V, find (i) Electric field strength: Use E=V/d and E=5000000 V/m
(ii) charge on the plates: Q=CV and Q= (iii) magnitude of electric flux density: Density will be charge per unit area, so: D=Q/A and D= >(d) Explain how Gauss’s theorem can be used to calculate the amount of charge on either plate and find its value when the voltage is as given in part (b).

So, = Charge density E is going to be made a constant, and integral of dA is just A.
So, we have: The As cancel and I'm left with the expression: How do I get charge from this?
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5 years ago
#2
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#3
Yes, but note that this value of doesn't coincide with the value obtained in (b)(ii).

Problem solved:
I was using the incorrect version of Gauss' law:
Should be using: Because this setup has a dielectric involved. The polarisation charge reduces the total charge inside or enclosed by the surface.

Working the above through, knowing that : Choosing a correct Gaussian surface will give: Where 0
X

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