### Author Topic: Capacitor with dielectric (constant Voltage and constant Charge mode)  (Read 33055 times)

#### Fu-Kwun Hwang

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• Posts: 3062
##### Capacitor with dielectric (constant Voltage and constant Charge mode)
« on: January 02, 2011, 11:31:23 pm »
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=   frac{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 dvec{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 \$Cequivfrac{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{kepsilon_0 (w-y)w}{d}\$
The total energy \$U=   frac{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{kepsilon_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).