Question

A parallel plate capacitor has plates of area A and separation d. We connect the capacitor to a battery of V volts. We then disconnect when the capacitor is fully charged.We now increase the plate separation by a small amount dz.What now is the energy stored in the capacitor? Express you answer in terms of the following variables if needed, A, V, d, dz and ϵ0 .

Answer 1

\[Q/V=Eo \times A \div d\] is the equation that relates everything before the plates are separated. When the plates are separated, Q remains the same so you could say \[Q = Eo \times A \times V / d\] and then plug that into the equation U = (1/2)QV. However, however, when the distance is changed, the voltage in the capacitor changes too. So \[Vnew = Q \times (d+dz) / (Eo \times A)\] then you can substitute in your previous equation for Q. Your final answer should be:
\[U = (1/2) (Eo \times A \times V \times dz/ d) (V + V/d)\] remember that the V in your answer is the V of the battery that charged the capacitor. Also, remember that when the plates are separated when the battery is disconnected, the charge remains constant and the voltage between the plates changes, but if the plates are separated when the capacitor is still connected to the battery, the voltage remains constant and the charge changes.