# Why is the energy stored in a capacitor E = 1/2QV?

the notes i found online say "Since the energy supplied by a power source is E = QV this means that capacitors store half the energy supplied to them (the rest is lost as heat in the wires and the power source)."

I understand the E = QV equation because P = IV and then multiplying each side by t gives E = QV.

But how can you know that exactly half the energy will be lost as heat in the wires and power source? I feel like the notes are simplifying it and that heat loss is not the real reason.

• Charge (Q) stored in a capacitor of capacitance (C) charged to a voltage (V) is given by:

Q = CV

In other words, if you plot a graph of Q against V you get a straight line, through the origin, of slope C.

If you look at your equation: E = QV and rearrange it:

V = E/Q

So in terms of the associated units:

volts = joules/coulombs.

Now ask what does the area under a graph of Q against V represent?

(coulombs) * (joules/coulombs) = joules

So what is the area of a triangle?

If you are into calculus:

E = ∫QdV = ∫CVdV = ½CV² + k

{E = 0,V =0 so k = 0}

• This is a problem in IMPEDANCE MATCHING.

Initially the volts across the capacitor are zero.

Yet the power supply has a value of V

So the current flowing into the capacitor is (v-0)/ R where R is whatever resistance is in the circuit.

R can be very small but it always exists.

The only difference caused by different values is the TIME taken to charge.

No matter what the time taken is then the average is still 1/2 V

so the energy supplied is VQ and the energy received is 1/2 VQ calculated over the whole duration of the charge.