Inserting a dielectric material between two plates of a capacitor

• idor
In summary, the energy stored in the capacitor is recovered through the decrease of the electric field between the capacitor plates.
idor
Homework Statement
Is the external mechanical work done by inserting a dielectric material between two plates of a capacitor (plates capacitor) when the power source remains connected always negative?
and if so, does it mean that the process can proceed spontaneously after being initiated?
Relevant Equations
The original capacity is C0, the voltage is V0, and the dielectric constant of the material is K >1.

1. new capacity: ##C = K\cdot C_{0}##.
2. initial potential energy: ##U_{i} = \frac 1 2 \cdot C_{0}V_{0}^2##.
3. final potential energy: ##U_{f} = \frac 1 2 \cdot KC_{0}V_{0}^2##.
4. total work done: ##W_{tot} =\Delta U = U_{f} - U_{i} = \frac 1 2 \cdot (K-1)C_{0}V_{0}^2 > 0##.
5. also total work done: ##W_{tot} = W_{p}~\text{(power source work for streaming more charge)} + W_{ext}~\text{(external work for inserting the material).}##
6. power source work: ##W_{p} = \Delta q \cdot V_{0} = (q_{f} - q_{i})V_{0} = (C \cdot V_{0} - C_{0} \cdot V_{0}) \cdot V_{0} = (K-1) \cdot C_{0}V_{0}^2.##
7. external mechanical work: ##W_{ext} = W_{tot} - W_{p} = \frac 1 2 \cdot (K-1) \cdot C_{0}V_{0}^2 - (K-1) \cdot C_{0}V_{0}^2 =
\\= -\frac 1 2 \cdot (K-1)C_{0}V_{0}^2 < 0.##

Last edited:

Well, you work out your own question immacculately ! For K > 1 you get a definite W < 0.
So: yes !
And it indeed means the capacitor 'pulls in' the dielectric. If the force is enough to overcome friction it will actually do it seemingly spontaneously.

So where does the energy come from ?

idor
Hi,
Thank you very much! your confirmation really helps.

It just seemed odd to me at first that the capacitor would 'pull in' the dielectric material even when the power source remains connected, after looking at a similar example where it was disconnected first, and the total amount of work done was negative because the capacitor lost potential energy.

However, in the second case (my question) the total amount of work done (positive from the power source + negative from the insertion of the material) is positive (as the capacitor gains potential energy), and that is why I wonder what makes this process favorable.

I assume that the energy comes from the power source, but what does the system earn from this?
Is it because the dielectric material lowers the electric field between the plates of the capacitor, and therefore somehow makes the system more stable?

Energy density (volumtric) is 1/2 DE. D = ##\epsilon E ##. So the final energy in the capacitor's E field is incremented by ## \epsilon_r = \epsilon/\epsilon_0. ##

Last edited:

1. What is a dielectric material?

A dielectric material is a type of insulating material that is placed between the two plates of a capacitor. It is usually made of non-conductive materials such as ceramic, plastic, or glass.

2. Why is a dielectric material used in a capacitor?

A dielectric material is used in a capacitor to increase the capacitance of the capacitor. This means that it can store more electric charge, making the capacitor more efficient.

3. How does a dielectric material affect the capacitance of a capacitor?

A dielectric material increases the capacitance of a capacitor because it reduces the electric field between the two plates, allowing more charge to be stored on the plates.

4. Can any dielectric material be used in a capacitor?

No, not all dielectric materials can be used in a capacitor. The material must have high dielectric strength and low dielectric loss in order to be effective in increasing the capacitance of the capacitor.

5. Can a dielectric material change the voltage of a capacitor?

No, a dielectric material does not change the voltage of a capacitor. It only affects the capacitance of the capacitor, which in turn affects the amount of charge it can store at a given voltage.

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