Work done when Inserting a Dielectric between Capacitor Plates

[lorx99]

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?

 

[kuruman]

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 $F = +\left( \frac{\partial U}{\partial x} \right)_V$

 

[lorx99]

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 :/.

 

[kuruman]

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

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."