Capacitor with dielectric (constant Voltage and constant Charge mode)

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).

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)

Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License

Please feel free to post your ideas about how to use the simulation for better teaching and learning.
Post questions to be asked to help students to think, to explore.
Upload worksheets as attached files to share with more users.

Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!

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).

Related Topics
	Subject	Started by	Replies	Views	Last post
	Constant acceleration in 1D x(t)/v(t) Kinematics	Fu-Kwun Hwang	0	23390	June 25, 2005, 10:04:48 am by Fu-Kwun Hwang
	Constant speed circular motion Kinematics	Fu-Kwun Hwang	3	17039	April 15, 2016, 11:19:25 pm by lookang
	vernier cliper (inch/mm mode, dial mode, measure inner/outer diameter and depth) misc	Fu-Kwun Hwang	11	96108	December 07, 2012, 10:49:32 pm by Fu-Kwun Hwang
	Constant acceleration in 1D x(t)/v(t) kinematics	ahmedelshfie	0	4303	May 27, 2010, 08:52:44 pm by ahmedelshfie
	charge particles attracted to a dielectric Request for physics Simulations	danielemateriale	3	4240	January 14, 2014, 09:02:24 am by AimeeSantana

	Author	Topic: Capacitor with dielectric (constant Voltage and constant Charge mode) (Read 32058 times)
0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).