Capacitors? Yes you CAN!

Reminder: The quizam will have a mixed bag of electrostatics AP problems, and might have another easy-to-predict surprise....

here's a hint:

12:41 PM, January 20, 2009

Solutions: Chapter 17, Q13, P32, 34, 37, 39, 47

Questions:

Problems:
32. Q = CV, so V = 22.0 volts.

34. using the equation for a parallel plate capacitor: C = e₀*A/d so the area = 5.0 x 10⁷ m^2. That's a lot of area!

37. We can get the electric field using V = Ed. Also, Q = CV.

We also need the fact that for a parallel plate capacitor, C = e₀*A/d.

We can put ALL of this together to get that Q = e₀*A*E*d/d = e₀*A*E , all of which we know! Substituting, Q = 2.63 x 10^-8 C.

39. Q = CV, so V = 90 volts. The potential difference V = Ed, so E = V/d = 4.5 x 10⁴ V/m or N/C.

47. The key to these problems is to use the equation in the form that does NOT change. For part (a), for example, it does NOT make sense to use U = 1/2*QV because the charge Q depends on the voltage through Q = CV. If you instead use U = 1/2CV², C is a constant, and V is the only thing that changes.

a) The energy must double, since U = 1/2CV² and C is a constant.

b) In this case, the Q is changed, and C must stay constant. We can then see by U = Q²/2C that the energy must be multiplied by a factor of 4.

c) If the battery stays connected, it is the VOLTAGE that stays the same, but the capacitance is divided by 2. Since U = 1/2CV^2, the energy will be divided by 2.