Work done when Inserting a Dielectric between Capacitor Plates

In summary: In the disconnected case the system is isolated and the insertion of the battery results in a reduction of the potential energy and a drop in the voltage across the capacitor. In the connected case, the battery pumps additional charge on the plates up to voltage V thereby increasing the potential energy of the system.
  • #1
lorx99
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Hi,

today in lecture, we discussed the work done on inserting a dielectric between a capacitor.

Two cases of this example:

One case, where the battery is disconnected so that the charge stays constant.

Other case where the battery stays connected so voltage is constant.

I am confused on why the work by the electric force for the disconnected battery is negative, while the work for the connected battery is positive. The professor discussed something about the fringed electric field that results in a force downwards by the electric field when we insert the dielectric. Also, he mentioned how when work is positive, energy is added, so it makes sense that the final energy state for inserting dielectric into a connected battery-capacitor increases. Vice versa to the disconnected battery..

I am confused on what is happening specifically for each case and more important "why".

Can someone please explain this concept?
 
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  • #2
The disconnected case is simple because there is only kind of electrical force that does work on the system and that is due to the electric field generated by the isolated charge on the plates. In that case the work done by the electrical forces is ##dW=(-dU)## and the force would be ##F = -\left( \frac{\partial U}{\partial x} \right)_Q## where ##x## is the direction of insertion and the derivative is explicitly taken at constant ##Q##.

The connected case is a bit more complicated because the work done by the battery needs to be taken into account. The work done on the system is ##dW=dW_{batt.}+(-dU)##. At constant voltage, when charge ##dQ## is added to the positive plate, ##dU=(1/2) V dQ##. Also, the work done by the battery to raise the potential energy of ##dQ## from zero to ##V## is ##dW_{batt}=VdQ##. Clearly then, ##dW_{batt}=2dU##. Thus the work done on the system at constant voltage is ##dW=2dU+(-dU)=+dU## and the associated force is ##F = +\left( \frac{\partial U}{\partial x} \right)_V##
 
  • #3
kuruman said:
The disconnected case is simple because there is only kind of electrical force that does work on the system and that is due to the electric field generated by the isolated charge on the plates. In that case the work done by the electrical forces is ##dW=(-dU)## and the force would be ##F = -\left( \frac{\partial U}{\partial x} \right)_Q## where ##x## is the direction of insertion and the derivative is explicitly taken at constant ##Q##.

The connected case is a bit more complicated because the work done by the battery needs to be taken into account. The work done on the system is ##dW=dW_{batt.}+(-dU)##. At constant voltage, when charge ##dQ## is added to the positive plate, ##dU=(1/2) V dQ##. Also, the work done by the battery to raise the potential energy of ##dQ## from zero to ##V## is ##dW_{batt}=VdQ##. Clearly then, ##dW_{batt}=2dU##. Thus the work done on the system at constant voltage is ##dW=2dU+(-dU)=+dU## and the associated force is ##F = +\left( \frac{\partial U}{\partial x} \right)_V##

Is there a more conceptual way to explain this like with the fringes? I just don't do well with equations/proofs that explain concepts :/.
 
  • #4
lorx99 said:
Is there a more conceptual way to explain this like with the fringes? I just don't do well with equations/proofs that explain concepts :/.
You asked for an explanation
lorx99 said:
on what is happening specifically for each case and more important "why".
Math is the language of physics and does a wonderful job when one wants to explain what specifically is going on with a system. If that's too much for you, you can wave your hands and say something like "In the disconnected case the system is isolated and the insertion of the battery results in a reduction of the potential energy and a drop in the voltage across the capacitor. In the connected case, the battery pumps additional charge on the plates up to voltage V thereby increasing the potential energy of the system."
 
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Related to Work done when Inserting a Dielectric between Capacitor Plates

1. What is the definition of "work done" when inserting a dielectric between capacitor plates?

Work done is the amount of energy expended or transferred when an external force is applied to displace an object. In the context of inserting a dielectric between capacitor plates, it refers to the energy required to move the dielectric material into the electric field between the plates.

2. How is the work done calculated in this scenario?

The work done when inserting a dielectric between capacitor plates can be calculated using the formula W = 0.5 x C x (V22 - V12), where W is the work done, C is the capacitance of the capacitor, V2 is the final voltage after inserting the dielectric, and V1 is the initial voltage before inserting the dielectric.

3. How does inserting a dielectric affect the work done in a capacitor?

Inserting a dielectric between capacitor plates increases the capacitance of the capacitor, which in turn decreases the voltage across the plates. As a result, the work done when inserting a dielectric is less than the work done without the dielectric, as the decrease in voltage leads to a decrease in the stored energy in the capacitor.

4. What happens to the work done if the dielectric material is removed from the capacitor?

If the dielectric material is removed from the capacitor, the capacitance of the capacitor decreases and the voltage across the plates increases. This results in an increase in the work done to remove the dielectric, as the stored energy in the capacitor increases.

5. Is the work done positive or negative when inserting a dielectric between capacitor plates?

The work done when inserting a dielectric between capacitor plates is negative. This is because the external force needed to insert the dielectric is in the opposite direction of the electric field, thus resulting in a negative value for the work done.

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