What is the value of C2 when charged by a battery and connected to C1?

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Homework Help Overview

The problem involves two capacitors, C1 and C2, where C1 is charged by a battery and then connected to an initially uncharged capacitor C2. The final voltage across both capacitors is given as 20V. The task is to determine the value of C2 based on the charge conservation principle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of charge and potential energy in capacitors, with attempts to calculate the value of C2 using different methods, including charge conservation and energy equations.

Discussion Status

Some participants have provided calculations and reasoning regarding the conservation of charge and energy. There is acknowledgment of different approaches, with one participant expressing clarity after reviewing another's method. However, no explicit consensus has been reached on the final value of C2.

Contextual Notes

Participants are working under the assumption that charge is conserved when the capacitors are connected. There is some uncertainty regarding the potential energy calculations and the initial conditions before C2 is connected.

PeachBanana
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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.
 
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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.
 
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
 
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
 
Ok, the way you have done it makes much more sense. Thank you.
 

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