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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?

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closed as off-topic by Rob Jeffries, Danu, Brandon Enright, Kyle Kanos, David Z Dec 23 '14 at 8:23

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  • $\begingroup$ That would be $\sigma A$, wouldn't it... Though your version of Gauss's law applies to both conduction and polarisation charges. $\endgroup$ – Rob Jeffries Dec 22 '14 at 22:37
  • $\begingroup$ @RobJeffries Assuming $\sigma=\frac{q}{A}$, then $q$ would indeed by $\sigma \times A$. But this gives me a different answer to my $Q=CV$ answer for the charge? $\endgroup$ – lmsavk Dec 22 '14 at 23:22
  • $\begingroup$ And that's why you need to use the correct version of Gauss's law! $\endgroup$ – Rob Jeffries Dec 23 '14 at 0:00
  • $\begingroup$ @RobJeffries I had no idea there were different versions? Can you point me toward the correct one? $\endgroup$ – lmsavk Dec 23 '14 at 0:37
  • $\begingroup$ Hint - when you cut through a dielectric, you are cutting through polarized atoms so some net charge will be on one side of your cut plane - no matter how you choose it. Ponder that... $\endgroup$ – Floris Dec 23 '14 at 2:43
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Your problem here is using the wrong version of Gauss's law, or at least one that is not applicable in this situation.

$$ \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$$ describes Gauss's law where $Q$ is the entire charge enclosed within the surface.

If we introduce a dielectric material, such that $\vec{D} = \epsilon_r \epsilon_0 \vec{E}$ and a polarisation field $\vec{P} = (\epsilon_r -1)\epsilon_0 \vec{E}$, then we can write $$ \vec{D} = \epsilon_0 \vec{E} + \vec{P}$$

If we take a closed surface integral of both sides $$ \oint \vec{D} \cdot d\vec{A} = \epsilon_0 \oint \vec{E} \cdot d\vec{A} + \oint \vec{P} \cdot d\vec{A}$$ But the right hand side is the total charge inside the surface (by Gauss's law) and the negative of the polarisation charge inside the surface - ie $Q - Q_p$. But the total charge $Q = Q_c +Q_p$, where $Q_c$ is the "conduction charge" - i.e. that due to freely moving charges. In other words, the polarisation charge reduces the total charge inside the surface and this is why you couldn;t get your solution.

Putting that all together, we get a new version of Gauss's law that is highly useful when there are polarisation charges about, which is that the closed surface integral of the D-field equals the total conduction charge inside the surface. $$\oint \vec{D} \cdot d\vec{A} = Q_c$$

That should do it.

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