Charging a Capacitor

  • #1

Homework Statement



A 4.6 mu or micro FF capacitor is charged to a potential difference of 15.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 4.6 mu or micro FF capacitor then drops to 7 V. What is the capacitance of the second capacitor?


Homework Equations



Q = CV

Capacitor in series
1/Ceq= 1/C1+1/C2


The Attempt at a Solution




I found the charge of the capacitor in the first scenario when the potential difference is 15 V

Q = CV
= 4.6E-6 F *15 V
= 7.35E-5 C

Since the potential drops to 7 V, you know the capacitors are in a series so that means the charge remains the same on both capacitors but potential changes

Capacitor in series
1/Ceq= 1/C1+1/C2

Q = Ceq * V

7.35E-5 C = (1/(4.6E-6 F) +1/C2)*7 V

I solved for C2...

C2 = -4.6E-6 C

- which is negative so its obvious that its wrong, but could someone tell me where I went wrong that would be greatly appreciated. Thank you.
 

Answers and Replies

  • #2
346
0
I think there is an error in your calculation of the charge on the first capacitor. It seems that your work is correct, but when I multiply (4.6 microF)*(15V) I get the charge to be
6.9e-5 C. Since you know that the voltage drops to 7V across the first capacitor, you can treat the circuit as if there is initially a 7V battery connected to only the second capacitor, which is unknown. In this case, you know that V=7V and Q= 6.9e-5C. From here you can manipulate the equation so that C=(Q/V). Therefore, C should equal 9.9 microF.
 
  • #3
Your way makes sense, and its a lot simpler than what I tried. The answer didn't work though so I'm going to try to keep double checking figures, i can't figure out what else would be wrong.
 
  • #4
152
0
Aren't the capacitors connected in parallel? And isn't the amount of charge calculated in the first part the total amount of charge in the 2-capacitor system in part two?
 
  • #5
152
0
clickcaptain,
I'm pretty sure you were doing it correctly in your first post (except for the first charge calculation, I got 6.9*10-5 C like w3390). Draw out the 2-capacitor circuit in the way it's described and I think you'll see that they are in parallel. When calculating C2, your Ceq should be Ceq = (C1 + C2) and not 1/Ceq = (1/C1) + (1/C2).

Are you arriving at the correct solution now?
 

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