AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Interpretation of the area under a graph of charge against pd.
Full specification wording
Energy stored by a capacitor (A-level only): Interpretation of the area under a graph of charge against pd.
For a capacitor, the charge Q stored is directly proportional to the potential difference V across it, so a graph of Q against V is a straight line through the origin with gradient equal to the…
Checked against the AQA 7408 specification (opens in a new tab) on .
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
For a capacitor, the charge Q stored is directly proportional to the potential difference V across it, so a graph of Q against V is a straight line through the origin with gradient equal to the capacitance C. The area under this graph between V = 0 and a chosen value V has the units of charge multiplied by potential difference, which is coulombs times volts, or joules. That area therefore represents the energy stored by the capacitor. Because the graph is a straight line, the area is the area of a triangle: one half times base times height, giving E = ½QV. Substituting Q = CV gives E = ½CV², and substituting V = Q/C gives E = ½Q²/C. These three forms are equivalent and each is useful depending on which quantities are known.
What earns the marks
- The graph of charge against potential difference is a straight line through the origin because Q = CV.
- The area under the graph has units of charge times potential difference, which are joules.
- The area under the graph represents the energy stored by the capacitor.
- For a straight-line graph through the origin, the area is ½ × base × height, giving E = ½QV.
- Substituting Q = CV or V = Q/C gives the equivalent forms E = ½CV² and E = ½Q²/C.
- The gradient of the graph is the capacitance C.
Common mistakes and how to put them right
Using the whole rectangle QV rather than the triangle ½QV.
Put it right: The area under a straight line through the origin is a triangle, so E = ½QV.
Treating the area as Q/V or V/Q.
Put it right: The gradient is C = Q/V, while the area is the energy.
Using E = ½QV when the graph is not a straight line through the origin.
Put it right: The formula applies to a capacitor obeying Q = CV.
Exam tips
- Sketch the Q against V graph and shade the triangular area to show clearly what is being calculated.
- State the units of the area as coulombs times volts, then convert to joules to justify the interpretation.
- Choose the energy form that matches the data given: use E = ½QV when Q and V are known, E = ½CV² when C and V are known, and E = ½Q²/C when C and Q are known.
Try the lesson
The start of the Energy stored by a capacitor (A-level only): Interpretation of the area under a graph of charge against pd. lesson, made from checked teaching notes for this specification point.
Have a go first
On a graph of charge Q against potential difference V for a capacitor, what does the area under the line represent?
Listen
For a capacitor, the charge Q stored is directly proportional to the potential difference V across it, so a graph of Q against V is a straight line through the origin with gradient equal to the capacitance C. The area under this graph between V = 0 and a chosen value V has the units of charge multiplied by potential difference, which is coulombs times volts, or joules. That area therefore represents the energy stored by the capacitor. Because the graph is a straight line, the area is the area of a triangle: one half times base times height, giving E = ½QV. Substituting Q = CV gives E = ½CV², and substituting V = Q/C gives E = ½Q²/C. These three forms are equivalent and each is useful depending on which quantities are known. A strong answer earns marks for these points. The graph of charge against potential difference is a straight line through the origin because Q = CV. The area under the graph has units of charge times potential difference, which are joules. The area under the graph represents the energy stored by the capacitor. For a straight-line graph through the origin, the area is ½ × base × height, giving E = ½QV. A common slip: Using the whole rectangle QV rather than the triangle ½QV: the error is forgetting the factor of one half. Instead, the area under a straight line through the origin is a triangle, so E = ½QV. Exam tip: Sketch the Q against V graph and shade the triangular area to show clearly what is being calculated.
Key words
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Graphical interpretation of area and gradient
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Equivalent forms of the energy equation
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Charge and potential difference in capacitors
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See every definition
- Graphical interpretation of area and gradient
- The area under a graph of charge against potential difference represents the energy stored, while the gradient is the capacitance C.
- Equivalent forms of the energy equation
- E = ½QV, E = ½CV² and E = ½Q²/C are equivalent, each useful depending on which quantities are known.
- Charge and potential difference in capacitors
- The charge Q stored is directly proportional to the potential difference V across it, so the graph is a straight line through the origin.
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