In simple terms
A friendly intro before the formal notes — no formulas yet.
Capacitor Energy: The Basics
Capacitors are like tiny rechargeable batteries, storing electrical energy in an electric field. The more charge and voltage, the more energy they hold, released gradually when discharged.
Think of a capacitor as a stretched spring or a compressed air tank. The more you stretch the spring or compress the air, the more potential energy it stores. Releasing it makes things move (or in our case, current flow).
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Charge Up: A capacitor collects charge on its plates, building up a potential difference.
- 2
Store Energy: This charge separation creates an electric field, storing electrical potential energy.
- 3
Calculate Power: Energy stored is proportional to charge and voltage, representable by the area under a Q-V graph.
- 4
Discharge: When connected to a circuit, the stored energy is released as current, decaying exponentially.
What this topic covers
The official Cambridge syllabus points this lesson works through.
- 19.2.1
Determine the electric potential energy stored in a capacitor from the area under the potential-charge graph
- 19.2.2
Recall and use
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Key formulas
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Tap a symbol — great for exam definitions
Tap a symbol — great for exam definitions
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Full topic notes
Formal explanation with the rigour you need for the exam.
Understanding Energy Storage
A capacitor stores energy by accumulating electric charge on its plates, creating a potential difference across them. This separation of charge establishes an electric field between the plates, and it's within this electric field that the electrical potential energy is held. The greater the charge and potential difference, the more energy is stored.
19.2 Energy stored in a capacitor.
The charge (Q) on a capacitor is directly proportional to its potential difference (V).
The area under the curve of a potential-charge graph is equal to the area under a triangle .
This area is the energy stored in a capacitor.
The energy stored (W) is therefore W =1/2 QV Substituting Q = CV we get W = ½ CV 2.
Calculating Stored Energy
The energy stored in a capacitor can be visualised and calculated by looking at its charge-voltage (Q-V) graph. Since capacitance (C) is constant, the relationship is linear, resulting in a straight line passing through the origin. The energy stored is simply the area of the triangle formed under this graph.
Using , we can derive:
Alternatively, using , we get:
Discharging a Capacitor: Releasing Energy
When a charged capacitor is connected to a resistor, it begins to discharge. The stored electrical potential energy is released, driving a current through the circuit. This process isn't instant; the charge, voltage, and current all decrease gradually following an exponential decay pattern over time. This decay is characterised by a specific time constant.
Energy is released as current flows during discharge.
Charge (Q), voltage (), and current (I) all decrease exponentially.
Discharge equations: and .
and are initial charge and voltage at .
The time constant () for an RC circuit is defined as .
is the time for Q, V, or I to fall to approximately 37% () of its initial value.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
A 220 . Calculate the energy stored. If it discharges through a 10 k\Omega resistor, what is the initial discharge current and the time constant of the circuit?
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Energy Stored (E):
A capacitor is charged to . Calculate the energy stored using .
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(2 s.f.)
How it all connects
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Glossary
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Quick check
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Revision flashcards
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In what form is energy primarily stored in a capacitor?
Electrical potential energy.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
- ✓
19.2 Energy stored in a capacitor.
- ✓
The charge (Q) on a capacitor is directly proportional to its potential difference (V).
- ✓
The area under the curve of a potential-charge graph is equal to the area under a triangle .
- ✓
This area is the energy stored in a capacitor.
- ✓
The energy stored (W) is therefore W =1/2 QV Substituting Q = CV we get W = ½ CV 2.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
9702/42 · Q6(b)(i)
Two capacitors P and Q are connected in parallel to a power supply of voltage V. The capacitance of P is 200 μF. The capacitance C_Q of Q can be varied between 0 and 400 μF. When C_Q = 0, the total energy stored in the capacitors is 2.5mJ. (i) Show that the supply voltage V is 5.0V.
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Checkpoint
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