# Find the capacitance from 2 energies stored, not knowing voltage

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1. Feb 28, 2015

### Steven Thomas

Hi there, new to this website so please bare with me if I break any rules and let me know for future.

I am a maths and physics tutor but occasionally I come across a question I just cannot work the answer for. In this case I know the answer but I can't get to it.

1. The problem statement, all variables and given/known data

Initially a charged capacitor stores 1600μJ of energy. When the pd across it decreases by 2.0V, the energy stored by it becomes 400μF. What is the capacitance of this capacitor?

A. 100μF
B. 200μF
C. 400μF
D. 600μF

2. Relevant equations
E=(CV2)/2
rearrange the above to give C=2E/V2

3. The attempt at a solution
So I've equated the two Cs to give me an equation that looks like E1/V2 = E2/(V-2)2 (the 2s have cancelled). I multiply each side by the opposite sides denominator and expand my squared bracket. Then by using the values of E given in the question I have a quadratic equation in terms of V that I can solve. This then gives me the values of V = 4 and V = 1.5. Using V = 4 with my E1 value I get the correct capacitance of 200μF. Do I know it must be V = 4 because I can't take 2volts away from V = 1.5?

Is there a better way for me to do this? This part of the exam is multi choice and the average amount of time meant to be spent on each question is 90 seconds, and this way takes a wee while. Any thoughts about a better way to solve it would be appreciated as when I tried to do it a different way (I've thrown the sheet of paper away with that working on and can't remember how I did it) I kept getting the capacitance as 400μF. But I have been given the answer of B, 200μF, and using this value with values of E given and V = 4 and V = 2 on hyperphysics' capacitance stored applet I know that it must be 200μF. I'm tearing my hair out with this and really could do with some help!

Thanks muchly,

Steven Thomas.

2. Feb 28, 2015

### Staff: Mentor

Whatever approach you take, I think you'll have the two ± possibilities to be investigated because of the square root involved.

3. Feb 28, 2015

### Steven Thomas

Surely there has to be a better way? Am I able to discard my second root of V=1.5 because 1.5-2 = -0.5 and I can't really have a negative voltage?

Steven Thomas.

4. Feb 28, 2015

### Staff: Mentor

A negative voltage would still represent energy stored. You'll need to test whether those voltages give the right energy levels, and if not, discard.

5. Feb 28, 2015

### Staff: Mentor

No. You know it can't be 1.5v because that voltage won't give you the 1600uJ you are needing here.

I calculate that C of 1800uF and voltages of 4/3 volt and -2/3 volt will also work. But 1800uF is not an available option in your multichoice answer.

6. Feb 28, 2015

### rude man

You could note that E varies as the square of V. So if V goes down by 2V then the original voltage must have been √(1600/400)*2V = 4V.
So simply C = 2E1/V12 = 3200/16 = 200 uF. There is no other answer possible.

7. Mar 9, 2015

### Steven Thomas

Thanks for your help here guys but I worked it out and it turned out to be real simple in the end.

I rearranged to make V the subject for both and then used the fact that E2 = E1/4 to give me a simultaneous set of equations I could solve real easy (also using V2=V1-2. Thanks anyway.