76 total
By the end of these notes, you should be able to:
When a capacitor charges up, a power supply (like a battery) pushes electrons from one plate to the other. This takes effort — the battery has to do work to move those charges. That work does not disappear; it gets stored in the capacitor as electric potential energy (energy held by charges due to their position in an electric field).
Think of it like pumping water uphill into a tank. The pump does work, and that work is stored as the potential energy of the water sitting up high. When you let the water flow back down, it can do useful work. A capacitor works the same way with electrical charge.
This is a really important idea to understand.
Because charge Q and potential difference V are directly proportional to each other (from Q = CV, if C stays constant, then V ∝ Q), the graph of potential difference (V) on the y-axis against charge (Q) on the x-axis is a straight line through the origin.
V (Volts)
| /
| /
| /
| /
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|/_______________
Q (Coulombs)
The gradient (slope) of this line equals 1/C (the reciprocal of capacitance).
Here is the key idea for Objective 1:
The electric potential energy stored in a capacitor is equal to the area under the potential–charge graph.
Because the graph is a straight line through the origin, the shape under the line is a triangle. The area of a triangle is:
Area = ½ × base × height
On this graph:
So the energy stored is:
W = ½ × Q × V
This is where the formula comes from — directly from the geometry of the graph. It is not just a rule to memorise; it has a clear, visual meaning.
Why ½ and not the full Q × V?
If you charged the capacitor at a constant voltage V the whole time, the energy would be Q × V. But the voltage starts at zero and rises gradually to its final value. So on average, the voltage during charging is only half the final voltage. That is why the energy is half of Q × V.
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