1. Energy and Capacitors
Remember that the voltage V is the work done per unit charge:
Imagine the capacitor is partially charged so that the charge on the plates is Q
It then acquires a little more charge δQ.
This involves the work of moving charge δQ from one plate to the other.
If δQ is very small V can be considered unchanged in which case
Q+ δQ Q + V - C

2. Energy and Capacitors And as Q+ δQ + + + + + V - - - - - C
We can substitute for V C
So the total work done in giving the capacitor full charge from 0 to Qfull
And in the limit as δQ→0

3. Writing Q fpr Qfull and making use of Q=VC
similarly
W =the energy stored by the charged capacitor (J)
Q= the charge on the plates (C)
V= the pd across the plates (V)
C = the capacitance of the capacitor (F)

4. Q=VC implies that graphs of Q against V must be straight line graphs
Energy Stored
Q=VC implies that graphs of Q against V must be straight line graphs
The gradient here is C Q
W =1/2 QV V

5. Energy Stored
The gradient here is 1/C V
W =1/2 QV Q

6. We know we have
The constant current means that we can work out the charge delivered from Q=It = 10 x 10-6 x 20 = 200 x 10-6C
Now apply