Class 12 Capacitance Practice: Battery Connected or Disconnected?
Before using Q=CV, decide what the circuit allows to change. A disconnected ideal capacitor keeps its charge; one still connected to an ideal constant-voltage source keeps its potential difference. The same dielectric can therefore lead to different charge and energy changes.
Curriculum checked 2 October 2026 · original practice for selected skills.
PREDICT · CHANGE · EXPLAIN
Fixed charge: change capacitance
An isolated ideal capacitor holds 24 μC. The model is V=24/C when C is entered in μF. There is no leakage.
The curve shows this simplified model over the labelled range. Read the exact point below; the sketch is not a measuring instrument.
Voltage: 6 V
Double C. Predict what happens to V before moving the slider.
Now close the explanation and try the independent retest. Moving a slider alone does not check your understanding.
Before you start
You will need: Potential difference, charge, micro-units and squared quantities.
Attempt each task before opening a hint. If you need help, reveal the first hint, then the setup, then compare your written steps with the solution. Finish with the retests without referring back.
Three ideas to keep in view
Capacitance measures charge stored per unit potential difference: C=Q/V. Microfarads and microcoulombs share the same factor of one millionth.
When a dielectric fully fills an ideal capacitor without changing its geometry, capacitance is multiplied by the relative permittivity. Check whether charge or voltage stays fixed.
Stored energy can be written as CV²/2 or Q²/(2C). Choose the form matching the fixed quantity rather than assuming energy always rises with capacitance.
Try the worked example first
P12-CAP-W1Disconnected capacitorFocused practice
A 4 μF capacitor is charged to 12 V and disconnected. A dielectric of relative permittivity 2 fully fills it. Find the new capacitance, charge, voltage and energy.
Show your working
Need a starting hint?
Disconnection prevents charge exchange in this ideal model.
Still stuck? Reveal the setup
First calculate Q=4 × 12 μC; then use the new capacitance.
Show worked solution+
1
Keep charge fixed
The original charge is 48 μC. New capacitance is 8 μF.
2
Find voltage and energy
Use Q/C and CV²/2 with the new values.
V=48/8=6 V,U=21(8)(62)=144μJ
Answer: 8 μF, 48 μC, 6 V and 144 μJ (initially 288 μJ).
Before moving on: can you explain why your method works, as well as give the answer?
Your turn: three different checks
Write a method as well as an answer. The tasks change the reasoning, not just the numbers.
P12-CAP-Q1Connected sourceFocused practice
A 3 μF capacitor remains connected to an ideal 10 V source while its capacitance doubles. Find the final charge.
Show your working
Need a starting hint?
The source maintains voltage, not charge.
Still stuck? Reveal the setup
Final capacitance is 6 μF.
Show worked solution+
1
Fix voltage
V remains 10 V.
2
Find charge
Q=CV gives 6 × 10 μC.
Answer: 60 μC
Before moving on: can you explain why your method works, as well as give the answer?
P12-CAP-Q2Series combinationFocused practice
Find the equivalent capacitance of 6 μF and 3 μF in series.
Show your working
Need a starting hint?
Reciprocals add in series.
Still stuck? Reveal the setup
1/C=1/6+1/3.
Show worked solution+
1
Add reciprocals
The sum is 1/2 per μF.
2
Invert
The equivalent capacitance is smaller than either individual capacitance.
Answer: 2 μF
Before moving on: can you explain why your method works, as well as give the answer?
P12-CAP-Q3Energy conversionFocused practice
An ideal 2 μF capacitor has 5 V across it. Find its stored energy in joules.
Show your working
Need a starting hint?
Use CV²/2 and convert μF to F.
Still stuck? Reveal the setup
C=2 × 10⁻⁶ F.
Show worked solution+
1
Square voltage
5²=25.
2
Calculate
Half of 2 × 10⁻⁶ × 25 is 25 × 10⁻⁶ J.
Answer: 25 μJ = 2.5 × 10⁻⁵ J
Before moving on: can you explain why your method works, as well as give the answer?
Catch a likely mistake
A tempting claim: “A dielectric always increases the capacitor's stored energy.”
Energy depends on whether voltage or charge is held fixed.
Close the examples and try again
These changed questions test whether you can reconstruct the method. If you use a hint, record where you got stuck and retry later. Completing this small set does not establish full chapter mastery.
P12-CAP-R1Changed dielectricIndependent retest
A 3 μF capacitor charged to 10 V is disconnected, then fully filled with dielectric of relative permittivity 3. Find final Q, V and energy.
Show your working
Need a starting hint?
Charge stays fixed.
Still stuck? Reveal the setup
New capacitance is 9 μF and Q is 30 μC.
Show worked solution+
1
Voltage
V=30/9=10/3 V.
2
Energy
Q²/(2C)=900/18 μJ.
Answer: 30 μC, 10/3 V, 50 μJ
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
P12-CAP-R2Changed connectionIndependent retest
Capacitors 2 μF and 5 μF are connected in parallel across 4 V. Find total charge.
Show your working
Need a starting hint?
Parallel capacitances add.
Still stuck? Reveal the setup
Ceq=7 μF.
Show worked solution+
1
Combine
Both capacitors share the same 4 V.
2
Find charge
Qtotal=7 × 4 μC.
Answer: 28 μC
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Choose the next useful step
If a retest exposed the same error, rewrite the first incorrect step and explain its correction aloud. If both were independent, return to a mixed exercise or a missed paper question. A parent can ask what changed in the method rather than only asking for the answer.
No. These questions use ideal capacitors, complete dielectric insertion and unchanged geometry. Real components may depart from those assumptions.
Can I print the questions and worked solutions separately?+
Yes. Use the two print buttons above. The question sheet leaves working space and hides solutions; the worked-solutions option includes the solution steps. Your browser can save either view as a PDF.
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