# Parallel Plates Capacitors, two capacitors.

1. Sep 22, 2012

### Freddd3

1. I mostly just need help understanding 3 and 4. And also understanding how to tell if things drawn on a circuit diagram/sketch are parallel or in series. I have the correct answer already, but it was mostly just guessing for 3 and 4.

The sketch is a side view of two capacitors consisting of parallel plates in air. The capacitor plates are equal in area but the plate separation differs as shown. Indivudual capacitors are specified with two letters, for example TR is a single capacitor. The charge on plate T is represented by Qt. The capacitors are charged so that the potential (voltage) at A, VA, initially equals 25 volts. for each of the statements choose the proper response.
Sketch here: http://www.freeimagehosting.net/m9tjl
http://www.freeimagehosting.net/m9tjl

1) ______ Qt + Qr is ... zero.
2) ______ The electric field between plates T and R is ... that between plates S and U.
3) ______ If the plate separation for capactior SU increases, the energy stored in TR will ... .
4) ______ If the plate separation for capacitor SU decreases, the charge on Qt will ... .
5) ______ Qt is ... Qs.
6) ______ The energy stored in capacitor SU is ... the energy stored in capacitor TR.

2. C = ε0 * A / d
C = Q/V
V = d*Q/(ε0 * A)
E = Q/ (ε0 * A)
σ = Q/A
U = (1/2) * C * V2

3. I have the correct answers for all of these questions, but I don't know why.

1.
EQUAL TO. This one is clear, because by definition parallel plate capacitors have charges with equal magnitudes, but opposite signs.

2.
LESS THAN. My reasoning was that the voltage for each one should be the same. So E = VA/d is how I figured out the electric field for each one. "d" is larger for TR than SU, and the voltage and the area for both SU and TR are the same. So the E is smaller in TR than SU.

3.
INCREASE. Don't know why. C gets smaller as d gets larger, and U gets smaller as C gets smaller. I figured if energy isn't leaving the system, then the energy from SU has to go somewhere, and so it must go to TR.

4.
DECREASE. C = Q/V. If voltage stays the same, then as C rises for SU (as D becomes smaller) then C must fall on the other side, and since voltage stays the same, then Q must be the thing that is falling in the equation C = Q/V.

5.
EQUAL TO. I figured You couldn't do the assignment if * something * didn't stay constant, so I assumed it must be the voltage. I'm guess that means that TR and SU are connected in parallel.

6.
LESS THAN. I used C = Q/V and C = ε0*A/d to figure it out. if V1 = V2 then if CTR was smaller than CSU, then QT must be smaller than QS

7.
GREATER THAN. U = . * C * V2 C was larger for SU than TR, and V1 = V2, so USU > UTR