Problem Types & Solved Examples
Six problem types. Every problem solved with examiner's intent, approach, and shortcut. This is the section where ranks are decided.
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All Types
Direct Formula
Conceptual
Multi-Step
Graph-Based
Assertion & Reason
Case-Based
Easy (CBSE)
Medium (NEET)
Hard (JEE)
Type 01
Direct Formula Problems
"Plug and chug. But know WHICH formula to plug."
P-001
A parallel plate capacitor has plates of area 0.02 m², separated by 2 mm. Find its capacitance. If a voltage of 100 V is applied, find the charge stored and energy stored.
Given Data
A = 0.02 m² | d = 2 mm = 2×10⁻³ m | V = 100 V | ε₀ = 8.85×10⁻¹² F/m
🔬 What examiner tests
Unit conversion (mm to m), correct formula application, and three-part calculation. CBSE awards 1 mark per correct calculation step.
1
Capacitance: C = ε₀A/d = (8.85×10⁻¹²×0.02)/(2×10⁻³) = 88.5 pF
2
Charge: Q = CV = 88.5×10⁻¹² × 100 = 8.85×10⁻⁹ C = 8.85 nC
3
Energy: U = ½CV² = ½×88.5×10⁻¹²×100² = 4.425×10⁻⁷ J = 0.4425 μJ
⚡ Quick check: U = ½QV = ½×8.85×10⁻⁹×100 = 4.425×10⁻⁷ J ✓ (Both methods give same answer)
A capacitor of capacitance 4 μF is charged to 12 V. Another capacitor of 6 μF is charged to 8 V. They are now connected in parallel (same polarity). Find: (a) common potential (b) energy loss.
Given
C₁=4μF, V₁=12V, C₂=6μF, V₂=8V (same polarity connection)
1
Common potential: V = (C₁V₁+C₂V₂)/(C₁+C₂) = (4×12+6×8)/(4+6) = (48+48)/10 = 9.6 V
2
Initial energy: Uᵢ = ½×4×10⁻⁶×144 + ½×6×10⁻⁶×64 = 288+192 = 480 μJ
3
Final energy: Uf = ½×(4+6)×10⁻⁶×9.6² = ½×10×10⁻⁶×92.16 = 460.8 μJ
4
Energy loss: ΔU = 480 - 460.8 = 19.2 μJ
⚡ Formula shortcut: ΔU = ½×(C₁C₂)/(C₁+C₂)×(V₁-V₂)² = ½×(24/10)×(12-8)² = ½×2.4×16 = 19.2 μJ ✓
Type 02
Conceptual Problems
"The answer is not in the formula. It's in your understanding."
A parallel plate capacitor is connected to a battery. A dielectric slab of dielectric constant K is then inserted between the plates. Which of the following quantities increases?
(A) Capacitance only (B) Capacitance and charge only (C) Capacitance, charge and energy (D) Only energy
(A) Capacitance only (B) Capacitance and charge only (C) Capacitance, charge and energy (D) Only energy
🔬 What examiner tests
Understanding of "battery connected" scenario. V remains constant. The trap: students say energy decreases because they confuse with battery-disconnected case.
🧠 Concept Selection
Battery connected → V = constant. Insert dielectric → C increases to KC₀. Since V constant: Q = CV → Q increases to KQ₀. Since U = ½CV² → U increases to KU₀. ALL THREE increase.
✓
Answer: (C) — Capacitance, charge, AND energy all increase by K.
❌ Why students choose (A)
They memorize that "dielectric increases capacitance" but forget that with battery connected, this forces more charge onto the plates and stores more energy (supplied by the battery).
Two capacitors C₁ = 3μF and C₂ = 6μF are connected in series to a 90V battery. Find: (a) charge on each capacitor, (b) voltage across C₁, (c) energy stored in C₂.
Given
C₁=3μF, C₂=6μF, V=90V (series connection)
🧠 Approach
Series → same charge on both. Find C_eq first → find total Q → find individual voltages.
1
C_eq = C₁C₂/(C₁+C₂) = 3×6/(3+6) = 18/9 = 2 μF
2
Q = C_eq × V = 2×10⁻⁶ × 90 = 180 μC (same on both in series)
3
V₁ = Q/C₁ = 180×10⁻⁶/(3×10⁻⁶) = 60 V
4
V₂ = 90 - 60 = 30 V (or Q/C₂ = 180/6 = 30 V) ✓
5
U(C₂) = ½×6×10⁻⁶×30² = ½×6×10⁻⁶×900 = 2.7 mJ
⚡ Voltage division in series: V₁/V₂ = C₂/C₁ = 6/3 = 2 → V₁=60V, V₂=30V. Check: 60+30=90V ✓
Type 03
Multi-Step Problems
"One wrong step anywhere = zero marks. Build the chain carefully."
Three capacitors C₁=2μF, C₂=3μF, C₃=6μF. C₁ is in series with the parallel combination of C₂ and C₃. A 120V battery is connected. Find: (a) equivalent capacitance, (b) total charge, (c) charge on C₁, (d) voltage across C₂, (e) energy stored in C₃.
Circuit Description
C₂ ∥ C₃ → then in series with C₁ → connected to 120V
🧠 Step-by-Step Framework
Always simplify from innermost combination outward. Here: first simplify C₂∥C₃, then put result in series with C₁.
1
C₂∥C₃: C₂₃ = 3+6 = 9 μF
2
C₁ series C₂₃: C_eq = (2×9)/(2+9) = 18/11 ≈ 1.636 μF
3
Total charge: Q_total = C_eq × V = (18/11)×10⁻⁶ × 120 = 2160/11 μC ≈ 196.4 μC
4
Charge on C₁: In series, C₁ gets full Q_total = 196.4 μC
5
V across C₁: V₁ = Q/C₁ = 196.4/2 = 98.2 V. So V across C₂₃ = 120-98.2 = 21.8 V (or Q/C₂₃)
6
V₂ = V₃: 21.8 V (parallel → same voltage)
7
Energy in C₃: U = ½×6×10⁻⁶×(21.8)² = 1.425 mJ
⚡ Reality check: Total energy = ½×C_eq×V² = ½×(18/11)×10⁻⁶×14400 = 11.8 mJ. Sum of U₁+U₂+U₃ should equal 11.8 mJ.
Type 04
Graph-Based Problems
"JEE loves graphs. Learn to read them, not just draw them."
A Q-V graph for a capacitor shows a straight line through origin with slope 4×10⁻⁶ C/V. The capacitor is charged to Q = 40 μC. Find: (a) capacitance, (b) voltage, (c) energy (from graph, without calculation), (d) what happens to the slope if plate separation is doubled?
🔬 What examiner tests
Graph interpretation (slope = C), ability to find energy as area under curve, and conceptual understanding of how geometry changes affect C.
1
Capacitance = Slope = 4×10⁻⁶ = 4 μF
2
V = Q/C = 40×10⁻⁶/(4×10⁻⁶) = 10 V
3
Energy from graph: = Area of triangle = ½×base×height = ½×10×40×10⁻⁶ = 200 μJ
4
If d doubles: C = ε₀A/d → C halves → slope becomes 2×10⁻⁶ C/V (less steep line)
🧠 Graph Insight
A steeper Q-V line = larger capacitance = more charge for same voltage. On the same graph, two capacitors with different slopes can be compared directly — the steeper one has higher capacitance.
Type 05
Assertion & Reason
"Both statements may be true — but is the reason correct?"
Assertion (A): When a charged capacitor is disconnected from the battery and then a dielectric slab is inserted, the energy stored in the capacitor decreases.
Reason (R): On inserting dielectric, the capacitance increases, and since charge remains constant, the energy Q²/2C decreases.
(A) Both A and R are correct, R is the correct explanation
(B) Both A and R are correct, R is NOT the correct explanation
(C) A is correct but R is wrong
(D) A is wrong but R is correct
Reason (R): On inserting dielectric, the capacitance increases, and since charge remains constant, the energy Q²/2C decreases.
(A) Both A and R are correct, R is the correct explanation
(B) Both A and R are correct, R is NOT the correct explanation
(C) A is correct but R is wrong
(D) A is wrong but R is correct
🧠 Analyze Each Statement
A: Battery disconnected → Q constant. Dielectric inserted → C increases. U = Q²/2C → U decreases. ✓ TRUE
R: The reason given is exactly correct — C increases with constant Q, so Q²/2C decreases. ✓ TRUE. And R correctly explains A.
R: The reason given is exactly correct — C increases with constant Q, so Q²/2C decreases. ✓ TRUE. And R correctly explains A.
✓
Answer: (A) — Both correct, R is the correct explanation.
🎯 A&R Strategy
First evaluate A alone. Then evaluate R alone. If both true, check if R logically proves A. Don't let a "true-sounding" R mislead you — it must specifically explain A using the correct mechanism.
Assertion (A): A capacitor cannot be used to store static charges in a permanent way without external connection.
Reason (R): The charges on the plates slowly leak through the dielectric medium, and the electric field between the plates prevents complete discharge.
Reason (R): The charges on the plates slowly leak through the dielectric medium, and the electric field between the plates prevents complete discharge.
🧠 Careful Analysis
A: TRUE — without connections, a real capacitor does slowly discharge due to leakage through imperfect dielectric.
R: Partially TRUE but WRONG explanation. Charges do leak, but the field does NOT prevent discharge — it actually drives leakage. The second part of R is incorrect.
R: Partially TRUE but WRONG explanation. Charges do leak, but the field does NOT prevent discharge — it actually drives leakage. The second part of R is incorrect.
✓
Answer: (B) — Both A and R have some truth, but R is NOT the correct explanation of A (because the field explanation is wrong).
Type 06
Case-Based Problems
"One scenario, multiple questions. Read carefully — context matters."
Passage: A parallel plate capacitor with plate area A = 10⁻² m² and separation d = 1 mm is connected to a 100V battery. A dielectric slab of thickness 0.5 mm and K = 2 is now inserted.
Case Questions
Q1: Find the original capacitance C₀
Q2: Find the new capacitance after inserting dielectric
Q3: What is the new charge on the capacitor?
Q4: By what factor does energy change?
Q2: Find the new capacitance after inserting dielectric
Q3: What is the new charge on the capacitor?
Q4: By what factor does energy change?
🧠 Note: Battery stays connected → V = 100V throughout
For partial dielectric insertion: treat as two capacitors in series — one air gap (d-t) and one dielectric gap (t).
1
C₀ = ε₀A/d = (8.85×10⁻¹²×10⁻²)/10⁻³ = 88.5 pF
2
New C = ε₀A/(d-t+t/K) = ε₀A/(0.5mm+0.5mm/2) = ε₀A/(0.75mm)
= (8.85×10⁻¹²×10⁻²)/(0.75×10⁻³) = 118 pF
= (8.85×10⁻¹²×10⁻²)/(0.75×10⁻³) = 118 pF
3
Q_new = C_new × V = 118×10⁻¹² × 100 = 11.8 nC
4
Energy ratio = C_new/C₀ = 118/88.5 ≈ 1.33× (energy increased by factor 4/3)
⚡ With battery connected: energy changes by same factor as capacitance. New C/Old C = d/(d-t+t/K) = 1/(0.75) = 4/3.