CIE AS/A Level Physics 19.2 Energy stored in a capacitor Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 19.2 Energy stored in a capacitor Study Notes – New Syllabus
CIE AS/A Level Physics 19.2 Energy stored in a capacitor Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- determine the electric potential energy stored in a capacitor from the area under the potential–charge graph
- recall and use W = 1/2 QV = 1/2 CV2
Electric Potential Energy from the Potential–Charge (V–Q) Graph
The energy stored in a capacitor can be found using the area under the potential–charge graph.
Key Idea:
The work done to charge a capacitor (and therefore the energy stored) is equal to the area under the \( \mathrm{V\!-\!Q} \) graph.
Because voltage increases linearly with charge for a capacitor (from \( \mathrm{V = 0} \) to \( \mathrm{V} \)), the graph is a straight line from the origin.
Energy stored = area of triangle under the V–Q line
\( \mathrm{E = \dfrac{1}{2}QV} \)
- The straight-line graph makes a right triangle.
- Base = \( \mathrm{Q} \), height = \( \mathrm{V} \).
- Area = \( \tfrac{1}{2} \times Q \times V \).
Using other energy formulas:
Using \( \mathrm{Q = CV} \), we get:
\( \mathrm{E = \dfrac{1}{2}CV^2} \)
\( \mathrm{E = \dfrac{Q^2}{2C}} \)
All three are equivalent.
Example
A capacitor is charged to \( \mathrm{Q = 4.0\times10^{-6}\ C} \) at a potential difference of 12 V. Find the energy stored using the area under the V–Q graph.
▶️ Answer / Explanation
Use energy formula from area:
\( \mathrm{E = \dfrac{1}{2}QV} \)
\( \mathrm{E = \dfrac{1}{2}(4.0\times10^{-6})(12)} \)
\( \mathrm{E = 2.4\times10^{-5}\ J} \)
Energy stored = \( \mathrm{2.4\times10^{-5}\ J} \)
Example
The V–Q graph for a capacitor shows that when the charge reaches \( \mathrm{6.0\times10^{-6}\ C} \), the potential difference is 15 V. Find the energy stored.
▶️ Answer / Explanation
Energy = area under straight-line graph:
\( \mathrm{E = \dfrac{1}{2}QV} \)
\( \mathrm{E = \dfrac{1}{2}(6.0\times10^{-6})(15)} \)
\( \mathrm{E = 4.5\times10^{-5}\ J} \)
Energy stored = \( \mathrm{4.5\times10^{-5}\ J} \)
Example
A capacitor has capacitance \( \mathrm{150\ \mu F} \). It is charged so its V–Q graph shows a straight line from (0, 0) to (Q, 50 V). Calculate:
- The charge \( \mathrm{Q} \)
- The energy stored
▶️ Answer / Explanation
Step 1: Find charge using \( \mathrm{Q = CV} \)
\( \mathrm{C = 150\ \mu F = 150\times10^{-6}\ F} \)
\( \mathrm{Q = CV = (150\times10^{-6})(50)} \)
\( \mathrm{Q = 7.5\times10^{-3}\ C} \)
Step 2: Energy from area under V–Q graph
\( \mathrm{E = \frac{1}{2}QV} \)
\( \mathrm{E = \frac{1}{2}(7.5\times10^{-3})(50)} \)
\( \mathrm{E = 0.1875\ J} \)
Energy stored = \( \mathrm{0.19\ J} \)
Energy Stored in a Capacitor: \( \mathrm{W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2} \)
The work done in charging a capacitor (and the energy stored in it) can be expressed using two equivalent formulas:
\( \mathrm{W = \dfrac{1}{2}QV} \) \( \mathrm{W = \dfrac{1}{2}CV^2} \)
- \( \mathrm{W} \) = energy stored (J)
- \( \mathrm{Q} \) = charge stored (C)
- \( \mathrm{V} \) = potential difference (V)
- \( \mathrm{C} \) = capacitance (F)
Why the factor \( \mathrm{\tfrac{1}{2}} \) appears:
As the capacitor charges, voltage increases from 0 to \( \mathrm{V} \), so the average voltage during charging is \( \mathrm{\tfrac{V}{2}} \). Multiplying by the final charge gives the total work done:
\( \mathrm{W = Q \left(\dfrac{V}{2}\right) = \dfrac{1}{2}QV} \)
Using \( \mathrm{Q = CV} \), we obtain:
\( \mathrm{W = \dfrac{1}{2}CV^2} \)
Example
A capacitor stores a charge of \( \mathrm{5.0\times10^{-6}\ C} \) at 10 V. Calculate the energy stored using \( \mathrm{W = \tfrac{1}{2}QV} \).
▶️ Answer / Explanation
\( \mathrm{W = \tfrac{1}{2}QV = \tfrac{1}{2}(5.0\times10^{-6})(10)} \)
\( \mathrm{W = 2.5\times10^{-5}\ J} \)
Energy stored = \( \mathrm{2.5\times10^{-5}\ J} \)
Example
A \( \mathrm{220\ \mu F} \) capacitor is charged to 25 V. Calculate the energy stored using \( \mathrm{W = \tfrac{1}{2}CV^2} \).
▶️ Answer / Explanation
Convert capacitance:
\( \mathrm{220\ \mu F = 220\times10^{-6}\ F} \)
\( \mathrm{W = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(220\times10^{-6})(25^2)} \)
\( \mathrm{W = 110\times10^{-6} \times 625} = 6.88\times10^{-2}\ J \)
Energy stored = \( \mathrm{6.9\times10^{-2}\ J} \)
Example
A capacitor is charged to 80 V and stores 0.32 J of energy. Calculate its capacitance using \( \mathrm{W = \tfrac{1}{2}CV^2} \), and then find the charge stored.
▶️ Answer / Explanation
Step 1: Find capacitance
\( \mathrm{W = \dfrac{1}{2}CV^2 \Rightarrow C = \dfrac{2W}{V^2}} \)
\( \mathrm{C = \dfrac{2(0.32)}{80^2}} \)
\( \mathrm{C = \dfrac{0.64}{6400} = 1.0\times10^{-4}\ F} \)
Step 2: Find charge using \( \mathrm{Q = CV} \)
\( \mathrm{Q = (1.0\times10^{-4})(80)} = 8.0\times10^{-3}\ C \)
Capacitance = \( \mathrm{1.0\times10^{-4}\ F} \)
Charge stored = \( \mathrm{8.0\times10^{-3}\ C} \)