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Electric Potential: Capacitor

  • #1
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Homework Statement



A 8.1 mu F capacitor is charged by a 155V battery (see the figure a) and then is disconnected from the battery. When this capacitor C_1 is then connected (see the figure b) to a second (initially uncharged) capacitor, C_2, the final voltage on each capacitor is 20V. What is the value of C_2?

Homework Equations



Q = CV

The Attempt at a Solution



(8.1*10^-6 F)(155 V) = 1.256 * 10^-3 C

1.256*10^-3 C / 20 V = 6.3*10^-5 F

I did it this way because I thought charge was conserved.
 

Answers and Replies

  • #2
CWatters
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I believe you have correctly calculated the combined capacitance (eg C1 and C2 in parallel). That's not quite what the question asked.
 
  • #3
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I tried to find the potential energy stored in the capacitors but I don't think I did it correctly.

I used PE = 1/2 Q*V

Q = 1.256 * 10^-3 C
V = 135 V (unsure)

PE = 0.0847 J

From there I tried to find the value of C_2 since it said charge is conserved by using:

C = .5*Q^2/PE

C = 7.4*10^-3 F
 
  • #4
CWatters
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If I remember correctly charge will be conserved (nowhere for it to go apart from being shared between the two capacitors).

So before C2 is connected..

Q = C1 * Vbefore

and after C2 is connected..

Q = (C1+C2) * Vafter

where

Vbefore = 155V
Vafter = 20V

equate

C1 * Vbefore = (C1+C2) * Vafter

Rearrange to give C2...

C1 * Vbefore = (C1 * Vafter) + (C2 * Vafter)

C2 * Vafter = C1 * Vbefore - (C1 * Vafter)

C2 = C1 (Vbefore - Vafter)/Vafter

Substitute values...

C2 = 8.1 * 10-6SUP] (155-20)/20

C2 = 54.7 * 10-6SUP] F

or 54.7 uF
 
  • #5
191
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Ok, the way you have done it makes much more sense. Thank you.
 

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