Can somebody tell my where I am going to wrong.

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The discussion focuses on deriving the formula W=1/2(V^2)C for work in a capacitor. The user is attempting to understand the derivation but is unsure about the steps involved. A key point made is the necessity of using integrals to account for the changing voltage as charge is added to the capacitor. The integral of V(q) multiplied by dq leads to the conclusion that W equals (1/2)CV^2. Understanding integrals is essential for completing this derivation accurately.
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W=work(energy) in joules
C=capacitance in farads
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
 
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dbernat32 said:
W=work(energy) in joules
C=capacitance in farads
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

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.
 
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