How Is Energy Calculated in a Capacitor?

[Cal124]

Homework Statement

Parallel plate capacitor, C = 1.7 uF
the difference between the two plates is 3 V
Find the energy stored in the capacitor

Homework Equations

U = 1/2 QV^2

The Attempt at a Solution

U = 1/2 (1.7x10^-6 F) (3V)^2 = 3.65x10^-6 J

Says the answer is: 1.53x10^-5 J

Not sure where I've gone wrong, any help would be great!

 

[haruspex]

You misquoted the equation... should be either QV/2 or CV²/2, but the start of the next line was ok.
I don't understand how you then got 3.65 etc. I think you should have got exactly half the given answer.
Are you sure the question is how much energy is stored, not how much energy was required to store the charge?

 

[Cal124]

"calculate the energy stored in the capacitor when the difference of potential between the two plates is 3V"
Apologies, yeah i meant CV²/2 but surly that is the same as 1/2 CV², so using the equation:
Capacitance = 1.7 uF = 1.7x10^-6 F
Difference of potential between the two plates = 3 V
// Energy stored = 1/2 . 1.7x10^-6 . 3²
// = 7.65x10^-6 J

Thanks for your help

 

[haruspex]

Apologies, yeah i meant CV²/2 but surly that is the same as 1/2 CV², so using the equation:

of course, but you wrote 1/2 QV².

 

[Cal124]

Any idea where I've gone wrong?

 

[haruspex]

Any idea where I've gone wrong?

I agree with your last answer.