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Gauss' Law Question

>A parallel plate capacitor with dielectric (as above), together with its dimensions. Its plates are square. The capacitance is given by the usual formula,

C = \frac{\epsilon _0 \epsilon _r A}{t}

Where:
A= 0.2

m^2

t= 0.1mm

\epsilon _r

= 2.2

\epsilon _0= 8.854 \times 10^{-12}

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=

1.95 \times 10^{-5}

(iii) magnitude of electric flux density: Density will be charge per unit area, so: D=Q/A and D=

9.75 \times 10^{-5} C/m^{2}

>(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,

\sigma

= Charge density

\oint E \cdot dA = \frac{Q_{enclosed}}{\epsilon _0}

E is going to be made a constant, and integral of dA is just A.
So, we have:

(E)(A) = \frac{\sigma A}{\epsilon _0}

The As cancel and I'm left with the expression:

E = \frac{\sigma}{\epsilon _0}

How do I get charge from this?

(edited 9 years ago)

Reply 1

uberteknik

Original post by halpme

\sigma

= Charge density

\oint E \cdot dA = \frac{Q_{enclosed}}{\epsilon _0}

E is going to be made a constant, and integral of dA is just A.
So, we have:

(E)(A) = \frac{\sigma A}{\epsilon _0}

The As cancel and I'm left with the expression:

E = \frac{\sigma}{\epsilon _0}

How do I get charge from this?

E =\frac{\sigma}{\epsilon_0}

also

\sigma = \frac{Q}{A}

therefore

E \epsilon_0 = \frac{Q}{A}

Q = EA \epsilon_0

Reply 2

halpme

Original post by uberteknik

E =\frac{\sigma}{\epsilon_0}

also

\sigma = \frac{Q}{A}

therefore

E \epsilon_0 = \frac{Q}{A}

Q = EA \epsilon_0

Yes, but note that this value of

Q

doesn't coincide with the value obtained in (b)(ii).

Problem solved:
I was using the incorrect version of Gauss' law:
Should be using:

\oint D\cdot dA = \epsilon _0 \oint E \cdot dA + \oint P \cdot dA

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

Q_{tot} = Q_c + Q_p

\oint D \cdot dA = Q_c

Choosing a correct Gaussian surface will give:

Q_c = DA

Where

D=\epsilon _0 \epsilon _r E

(edited 9 years ago)

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