Question

Electromagnetism question (file attached)

Answer 1

So, it's question b) that I don't quite get how to solve.

Answer 2

The plates are oppositely charged so should attract before anything else is considered. So **negative** work will be required to bring them together.

You can tackle this a bunch of ways, and looking at the energy released \(\Delta U\) when the are brought together is one way. That will be the negative work done: \(W = -\Delta U\)


I assume you know the energy stored in a cap is

\(U = \dfrac{1}{2} CV^2 = \dfrac{Q^2}{2C} \qquad \triangle\)

And the capacitance of a parallel plate is \(C = \dfrac{\varepsilon A}{x}\) , for separation \(x\).

**critically**, Q is fixed/ conserved as the cap is removed from the source so we should go with

\(U = \dfrac{Q^2}{2C} = \dfrac{Q^2 x}{2 \varepsilon A}\) so that \(\Delta U \propto \Delta x\)

so you'd have to do work on the plates to further separate them , and the energy stored would increase.....

this is from the defintion \(C = \dfrac{Q}{V}\) and a bunch of stuff i reckon you will already know how to derive...including the energy stored in a cap.

and you can see what would happen to the potential drop across the cap as the plates are separated by having a further play with \(\triangle\).

make sense ?!

Answer 3

Thank you so much @IrishBoy123 That makes sense, yes:-)

Answer 4

cheers mate!