1
$\begingroup$

Consider a parallel plate capacitor having capacitance $ C$ and charge $Q$. then the energy of the capacitor will be $\frac{Q^2}{2C}$. Now if a dielectric is inserted, then $C $ increases and thus the energy decreases.

My question is - where does the energy go? If it gets used up to polarise the dielectric, thne if we remove the dielectric, how does it regain the energy from the dielectric?

$\endgroup$
1
$\begingroup$

My question is - where does the energy go?

Someone has to insert or remove the dielectric and that will require work (either positive or negative depending on the situation). Work is where the energy goes.

$\endgroup$
  • $\begingroup$ More specifically, the person who inserts the dielectric will have to stop it from oscillating (it is sucked in and will hence overshoot). Inserting it does not require work, removing it does (your sign is wrong). Otherwise this is a perfect answer. $\endgroup$ – pyramids Mar 7 '15 at 9:17
  • $\begingroup$ I think that inserting can also require work on the inserters behalf... if what is already in there is more dielectric that what is being inserted. E.g., an oil-filled capacitor gap into which foam is inserted... besides work can be positive or negative. $\endgroup$ – hft Mar 7 '15 at 17:47
  • $\begingroup$ Of course, yes. Unless you read the question to imply that there must not have been a dielectricum inside the capacitor yet, as I did. $\endgroup$ – pyramids Mar 7 '15 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.