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

In summary, a 8.1 mu F capacitor charged by a 155V battery and then connected to a second uncharged capacitor resulted in a final voltage of 20V on both capacitors. Using the equation Q = CV, the value of C2 can be calculated as 6.3*10^-5 F. However, in order to calculate C2 using the conservation of charge, the equation C2 = C1 (Vbefore - Vafter)/Vafter can be used, resulting in a value of 54.7 uF.
  • #1
PeachBanana
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0

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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  • #2
I believe you have correctly calculated the combined capacitance (eg C1 and C2 in parallel). That's not quite what the question asked.
 
  • #3
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
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
Ok, the way you have done it makes much more sense. Thank you.
 

1. What is electric potential?

Electric potential is a measure of the electric potential energy per unit charge at a specific point in an electric field. It is also known as voltage.

2. How is electric potential related to a capacitor?

A capacitor is a device that stores electric potential energy in an electric field. The electric potential across a capacitor is directly proportional to the amount of charge stored on it.

3. What is the formula for calculating the electric potential of a capacitor?

The formula for calculating the electric potential of a capacitor is V = Q/C, where V is the voltage, Q is the charge stored on the capacitor, and C is the capacitance of the capacitor.

4. How does the distance between the plates of a capacitor affect the electric potential?

The distance between the plates of a capacitor affects the electric potential by changing the strength of the electric field. The closer the plates are, the stronger the electric field and therefore the higher the electric potential.

5. What is the unit of measurement for electric potential?

The unit of measurement for electric potential is the volt (V). One volt is equivalent to one joule of energy per coulomb of charge.

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