Two parallel plates, each having area A = 3557 cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.59 cm. The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.18 cm). What is the energy stored in this new capacitor?

Question

jamesj · Answer

Well, Capacitance is given by
$$ C = \frac{ \epsilon A}{d} $$
where $ A $ is the area of the plates, $ d $ the distance between them.  
And the work done in creating a charged capacitor, i.e., the energy stored by them is
$$ W = \frac{1}{2} CV^2 $$

Now find the value of W for the final state of your system.

anonymous · Answer

Doesn't the voltage change when you remove the battery?

jamesj · Answer

No, that's the point.  The plates are charged by the battery, and then when you take the battery away (and do NOT close the circuit, of course!), the plates stay charged.

anonymous · Answer

Thanks.

anonymous · Answer

Wait but looking at my textbook it says $$U=(1/2) QV$$. If neither voltage or Q change then the energy stored would be the same.

jamesj · Answer

No, that's not right. Work must be done. Watch the first five minutes of this lecture on capacitors: http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-7-capacitance-and-field-energy/

anonymous · Answer

So does V change when you increase the distance? Cause thats the only thing I can think of that might explain the confusion.

anonymous · Answer

For the sake of closure I will post what I think the solution is: V does change with the change in distance. The new voltage is 12V and stored energy (U) is 1.92*10^-8 J.

jamesj · Answer

Yes, V, does change.  Now $ C = Q/V $ is the fundamental definition of capacitance and thus
$$ U = \frac{1}{2}CV^2 = \frac{1}{2} \frac{Q^2}{C} $$