How Much Energy Does a Battery Supply to Charge Parallel Capacitors?

[physgrl]

Homework Statement

Two capacitors having values of 10 μF each are connected in parallel and then hooked up to a 9.0 V battery. How much energy must the battery supply to charge the capacitors?

*a. 8.1 x 10-4 J
b. 9.0 x 10-5 J
c. 1.6 x 10-3 J
d. 2.0 x 10-4 J
e. 2.3 x 10-5 J

Homework Equations

ε=ΔW/ΔQ
Q=CV
ΔW=ΔE
C=C1+C2...+Cn (for capacitors in parallel)

The Attempt at a Solution

ΔE=ε*ΔQ
ΔE=ε*CV
ΔE=9V*(10μF+10μF)*9V
ΔE=1.62x10^-3 J

the answer key says A is correct...can someone tell my why?

 

[gneill]

Your calculation made the assumption that the change in potential is the same for all the charges moved onto the capacitor. When an initially uncharged capacitor is connected to a voltage source for charging up, the potential across the capacitor is not immediately equal to that of the source; the capacitor begins with 0V across it which rises to meet the source's potential as current flows and its charge increases.

What this means is that it takes less work to move charge onto the capacitor early on in the process when the capacitor potential is low than it does later when the potential is higher. If you perform the required integration to calculate the work you'll find that the resulting expression is similar in form to other such expressions for other types of energy. This can make it easier to remember them all! Some examples:

Spring potential Energy: PE = (1/2)kx²
Linear Kinetic Energy: KE = (1/2)mv²
Rotational Kinetic Energy: KE = (1/2)Iω²
Capacitor Electrical PE: PE = (1/2)CV²

 

[technician]

You had an almostidentical question to this a couple of weeks ago. See the first post on the list at the bottom.
There are 2 important pieces of information
1) The energy STORED on the capacitor is (a) 8.1 x 10^-4 J
2) The energy supplied by the battery is (c) 1.62 x 10^-3 J
Whichever answer you take to be correct depends on your interpretation of the question.
For me, this means it is a badly worded question. A multiple choice question must have only one correct answer with no doubt whatsoever.

 

[physgrl]

OK...thanks to the both of you!

 

[asteriatic]

The correct answer is A, 8.1 x 10-4 J.

Your calculation for the total capacitance is incorrect. When capacitors are connected in parallel, their capacitances add up, but not in a simple addition like you have done. It should be C = C1 + C2 = 10 μF + 10 μF = 20 μF.

Using the correct value for capacitance, the calculation becomes:

ΔE = 9V * 20μF = 180 μJ = 8.1 x 10-4 J

Therefore, the battery must supply 8.1 x 10-4 J of energy to charge the capacitors.