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Electrostatic Potential Top Capacitance Numericals Power of Capacitors
Capacitance is one of the most scoring chapters in Physics—especially in board exams. Most questions are formula-based, predictable, and easy to solve if your concepts are clear.
In this guide, we’ll cover important formulas, step-by-step numerical solving methods, common mistakes, and exam strategies.
Capacitance is the ability of a conductor to store charge.
Formula: C = Q / V
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| Concept | Formula | When to Use |
|---|---|---|
| Basic Capacitance | C = Q / V | Direct definition |
| Parallel Plate Capacitor | C = ε₀ A / d | Two plates in vacuum |
| With Dielectric | C = K ε₀ A / d | Medium inserted |
| Energy Stored | U = 1/2 CV² | Energy calculations |
| Series Combination | 1/C = 1/C₁ + 1/C₂ | Capacitors in series |
| Parallel Combination | C = C₁ + C₂ | Capacitors in parallel |
| Electrical Quantity | Water Analogy |
|---|---|
| Charge (Q) | Water stored |
| Voltage (V) | Water pressure |
| Capacitance (C) | Tank size |
Key Idea: Bigger tank → more water stored → higher capacitance
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Question: A capacitor stores 6 μC charge at a potential of 3V. Find capacitance.
Solution:
C = Q / V = (6 × 10⁻⁶) / 3 = 2 × 10⁻⁶ F
Final Answer: 2 μF
Question: Find capacitance if area = 2 m² and distance = 0.01 m.
Solution:
C = ε₀ A / d
C = 8.85 × 10⁻¹² × (2 / 0.01)
C = 1.77 × 10⁻⁹ F
Final Answer: 1.77 nF
Question: A dielectric of constant 5 is inserted. Find new capacitance if original is 2 μF.
Solution:
C' = K × C = 5 × 2 = 10 μF
Final Answer: 10 μF
| Situation | What Changes | What Remains Constant |
|---|---|---|
| Battery connected | Charge changes | Voltage constant |
| Battery removed | Voltage changes | Charge constant |
| Dielectric inserted | Capacitance increases | Depends on condition |
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Question: Two capacitors 4μF and 6μF are in series. Find equivalent capacitance.
Solution:
1/C = 1/4 + 1/6
1/C = (3 + 2) / 12 = 5/12
C = 12/5 = 2.4 μF
Final Answer: 2.4 μF
Question: Capacitors 3μF and 5μF are in parallel. Find total capacitance.
Solution:
C = 3 + 5 = 8 μF
Final Answer: 8 μF
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| Property | Series | Parallel |
|---|---|---|
| Capacitance | Decreases | Increases |
| Charge | Same | Different |
| Voltage | Divides | Same |
| Formula | Reciprocal sum | Direct sum |
Question: Find energy stored in a capacitor where C = 2μF and V = 10V.
Solution:
U = 1/2 CV²
U = 1/2 × 2 × 10⁻⁶ × 100
U = 1 × 10⁻⁴ J
Final Answer: 0.0001 J
If voltage doubles, energy becomes 4 times because energy is proportional to V².
Shortcut:
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Problem: Students confuse series and parallel formulas.
Agitate: This leads to wrong answers and loss of marks in exams.
Solution: Series = Smaller capacitance Parallel = Larger capacitance
| Mistake | Correct Approach |
|---|---|
| Forgetting units | Always convert μF to F |
| Wrong substitution | Write formula first |
| Skipping steps | Solve step-by-step |
| Mixing formulas | Identify question type first |
Question: A 4μF capacitor is connected to a 12V battery. Battery is removed and dielectric (K=2) is inserted. Find new energy.
Solution:
Initial Energy:
U = 1/2 × 4 × 10⁻⁶ × 144 = 2.88 × 10⁻⁴ J
After dielectric (battery removed):
U' = U / K = (2.88 × 10⁻⁴) / 2 = 1.44 × 10⁻⁴ J
Final Answer: 1.44 × 10⁻⁴ J
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| Step | What to Do |
|---|---|
| 1 | Identify question type |
| 2 | Write correct formula |
| 3 | Convert units |
| 4 | Solve step-by-step |
| 5 | Highlight final answer |
Don’t just read—solve at least 20 numericals to improve speed and accuracy.
Download Free Notes & Practice Questions
| Question Type | Formula |
|---|---|
| Charge & Voltage | C = Q / V |
| Parallel Plate | C = ε₀ A / d |
| Energy | U = 1/2 CV² |
| Series | Reciprocal formula |
| Parallel | Direct addition |
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Capacitance is one of the highest scoring topics in Physics. With clear concepts, regular practice, and avoiding common mistakes, you can easily score full marks.
Focus on numericals, stay consistent, and you’ll master this chapter.
If you want to practice this topic, you can take a quiz in Curious Corner for better practice.
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