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 Author Topic: Capacitor with dielectric (constant Voltage and constant Charge mode)  (Read 30757 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: January 02, 2011, 10:31:23 pm » posted from:Taipei,T'ai-pei,Taiwan

A object with dielectric constant k is insert in between a capacitor with separation d, area A(=w*w).
If the capacitor is connected to a constant voltage source (Voltage =V),
The object will oscillate if it was not fit right into the capacitor.

Click play to start the simulation.

While in paused mode, you can adjust k,d => so capacitance C is changed,too.
You can watch how charge Q changed as C is changed in constant voltage mode.

If you turn into constant Q mode (switch off the charging circuit by selecting constant Q radio button),
the change Q will become constant,
and voltage V will vary if you move the dielectric object with slider (change y).

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The energy stored in the capacitor C can be expressed as $U=\tfrac{1}{2}CV^2$ when the capacitor are connected to a constant voltage source.
For two parallel plate capacitor with area A and separation d:
From Gauss law: $\int \vec{E}\cdot d\vec{s}=\frac{Q}{\epsilon_0}$
The electric field $E=\frac{\sigma}{\epsilon_0}$ where $\sigma$ is the charge density.
Since $V=E d=\frac{\sigma}{\epsilon_0} d =\frac{\sigma A d}{\epsilon_0 A}=\frac{Q d}{\epsilon_0 A}$,
so $C\equiv\frac{Q}{V}=\frac{\epsilon_0 A}{d}$.

If an dielectric material with dielectric constant k is insert between the parallel plate,
electric field will introduce dipole field (in opposite direction to electric field) inside the dielectric material.
So that the net elctric field become $E'=E/k$ (from defition of dielectric constant).
The voltage between capacitor plate will reduced to $V'=V/k$ if the capacitor is not connected to voltage source.
(The charge Q remain the same --- constant Q mode)

However, if the capacitor is connected to constant voltage source, then voltage source will charge the capacitor to keep up the same voltage V (from V' to V).
So more charge (Q'=kQ )will be stored into the capacitor, so the capacitance $(C'=kC)$.

If the dielectric meterial is moved side way, it will be attracted back to it's original position.
assume the side way displacement is y, and area $A=w*w$.
The capacitoance become $C''=\frac{\epsilon_0 A_1}{d}+\frac{\epsilon_0 A_2}{d}=\frac{\epsilon_0 yw}{d}+\frac{k\epsilon_0 (w-y)w}{d}$
The total energy $U=\tfrac{1}{2}C'' V^2$, so the force can be calculated as
$F_y=-\frac{dU}{dy}=-\frac{1}{2} \frac{dC''}{dy}V^2 =-\frac{1}{2}(\frac{\epsilon_0 w}{d}+\frac{k\epsilon_0 (-w)}{d})V^2=- (k-1) \frac{\epsilon_0 w}{2d}V^2$ (it is a constant acceleration, but in the direction opposite to displacement).
 E_capacitor_dielectri.jpg (15.54 KB, 398x448 - viewed 1287 times.) Logged
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