# Can somebody tell my where I am going to wrong.

1. Jan 23, 2008

### dbernat32

W=work(energy) in joules
A=Amperes
V=Voltage
Q=charge in coulombs
s=seconds
P=power in watts

I know that W=1/2(V^2)C is the correct formula, but I don't know how to derive it.

I am doing the following: C=Q/V, C=As/V, VC=VAs/V, VC=Ps/V, VC=W/V, W=C(V^2)
What am i missing?

--dbernat32

2. Jan 23, 2008

### lightarrow

Don't know if you studied integrals.
When you add a small charge dq to a condenser at potential V and charge q, you have to make the work V*dq. But now V is not the same anylonger because V = q/C and q is now different. So you have to write V(q) and dW = V(q)*dq.
To get the total work you have to sum all these infinitesimal quantities, that is you have to compute the integral:
Integral(0;Q) V(q)*dq = Integral(0;Q) (q/c)*dq = (1/2)Q^2/C = (1/2)CV^2.
The equality coloured in blue requires knowledge of integrals.

Last edited: Jan 23, 2008