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Intermediate

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Capacitors, Capacitance & Dielectrics

Master Capacitors Capacitance and Dielectrics for JEE Main Advanced NEET and CBSE Class 12 Physics. Covers definition C=Q/V geometry dependence, parallel plate capacitor ε₀A/d with dielectric Kε₀A/d, spherical and cylindrical capacitor formulas, series combination 1/Cs=Σ1/Ci same charge, parallel combination Cp=ΣCi same voltage, charge sharing V=Q/C series worked example, effect of dielectric at constant voltage C×K Q×K U×K E unchanged, at constant charge C×K V÷K U÷K E÷K energy paradox, charge redistribution common voltage formula, with fully verified JEE NEET board-level practice questions.

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A capacitor is the electrical analog of a spring — it stores energy in an electric field, releases it on demand, and its "stiffness" (capacitance) determines how much charge it stores per volt. From the flash in your camera to the timing circuits in your phone, capacitors are everywhere. In physics, the parallel plate capacitor is the foundational model that makes the concept concrete. Understanding how capacitance is calculated, how capacitors combine in series and parallel, how a dielectric changes the capacitance, and where the energy is stored — these are the core skills tested in JEE and NEET. Capacitor problems appear in almost every competitive exam paper.

1. Capacitance — Definition

A capacitor is a device that stores charge (and hence energy) by maintaining a potential difference between two conductors. The Capacitance C is the charge stored per unit potential difference:

$C = \frac{Q}{V}$

SI unit: Farad (F) = C/V. Practical units: μF (10⁻⁶ F), nF (10⁻⁹ F), pF (10⁻¹² F).

Capacitance depends only on the geometry of the conductors and the medium between them — NOT on Q or V.

2. Parallel Plate Capacitor

Two large parallel conducting plates, each of area A, separated by distance d (d ≪ √A). E = σ/ε₀ between the plates, V = Ed.

$C = \frac{ε_{0} A}{d}$

With dielectric (κ or K inserted between plates):

$C = \frac{K ε_{0} A}{d}$

K = dielectric constant (relative permittivity) ≥ 1. Increases C by factor K.

Other Capacitor Geometries

Type	Formula
Spherical (radius R, isolated)	C = 4πε₀R
Spherical (inner R₁, outer R₂)	C = 4πε₀R₁R₂/(R₂−R₁)
Cylindrical (length L, radii a,b)	C = 2πε₀L/ln(b/a)

3. Combination of Capacitors

Series Combination

Same charge Q on each; voltages add: V = V₁ + V₂ + V₃

$\frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + \dots$

C_series < smallest individual capacitor.

Parallel Combination

Same voltage V across each; charges add: Q = Q₁ + Q₂ + Q₃

$C_{p} = C_{1} + C_{2} + C_{3} + \dots$

C_parallel > largest individual capacitor.

Worked Example

C₁ = 4 μF, C₂ = 6 μF connected in series to V = 100 V.

Cs = C₁C₂/(C₁+C₂) = 24/10 = 2.4 μF

Q = CsV = 2.4×10⁻⁶ × 100 = 240 μC (same on both)

V₁ = Q/C₁ = 240/4 = 60 V; V₂ = 240/6 = 40 V; V₁ + V₂ = 100 V ✓

4. Energy Stored in a Capacitor

$U = \frac{1}{2} C V^{2} = \frac{Q^{2}}{2 C} = \frac{1}{2} Q V$

The energy is stored in the electric field between the plates. Energy density (energy per unit volume):

$u = \frac{U}{A d} = \frac{1}{2} ε_{0} E^{2}$

This energy density formula is universal — applies to any electric field, not just capacitors.

5. Effect of Dielectric

A dielectric is an insulating material whose molecules polarise in an electric field, creating an opposing internal field that reduces the net field inside. The dielectric constant K (= ε_r) is the ratio of the original field to the reduced field.

Quantity	Battery connected (V constant)	Battery disconnected (Q constant)
Capacitance C	Increases by K (KC)	Increases by K (KC)
Charge Q	Increases by K (more from battery)	Unchanged (no battery)
Voltage V	Unchanged (battery maintains V)	Decreases by K (V/K)
Electric field E	Unchanged (E = V/d)	Decreases by K (E/K)
Energy U	Increases by K (KU)	Decreases by K (U/K)

Capacitor Quick Reference

Scenario	Key Formula / Fact
Series: same ___ on each	Charge Q same; 1/Cs = Σ1/Ci
Parallel: same ___ on each	Voltage V same; Cp = ΣCi
Two capacitors in series	Cs = C₁C₂/(C₁+C₂)
Energy stored	½CV² = Q²/2C = ½QV
Dielectric at constant Q	C×K; V÷K; U÷K (energy released → dielectric pulled in)
Dielectric at constant V	C×K; Q×K; U×K (battery supplies extra energy)

JEE/NEET Power Tips

Series: same Q; Parallel: same V. This single mnemonic covers 80% of combination problems. In series, the middle capacitors are isolated — they must carry equal and opposite charges → same Q. In parallel, both plates connect to the same terminals → same V.
Charge redistribution when capacitors are connected: When a charged capacitor is connected to an uncharged one (in parallel), total charge redistributes: Q_total = Q_initial, V_common = Q_total/(C₁+C₂). Energy is always lost (to heat/radiation) unless the capacitors had the same voltage.
When dielectric is inserted with battery connected: energy INCREASES (battery supplies more charge). When inserted after battery disconnected: energy DECREASES (dielectric is pulled in, doing work). The energy paradox is a classic JEE question.
JEE ADV Partial dielectric insertion: Treat as two capacitors in parallel (dielectric region + air region), each with area A₁ and A₂ where A₁ + A₂ = A total.

Practice Questions

Q1 (JEE Main): Three capacitors of 2 μF, 3 μF, and 6 μF are connected in series. Find the equivalent capacitance and the charge on each when connected to 12 V.

1/Cs = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 → Cs = 1 μF

Q = CsV = 1×10⁻⁶ × 12 = 12 μC (same on all three in series)

V₁ = 12/2 = 6V; V₂ = 12/3 = 4V; V₃ = 12/6 = 2V. Sum = 12V ✓

Q2 (NEET): A parallel plate capacitor of capacitance 10 μF is charged to 200 V. Find the energy stored. If a dielectric of K = 4 is now inserted with the battery still connected, find the new energy.

U_initial = ½CV² = ½ × 10×10⁻⁶ × 200² = 0.2 J = 200 mJ

Battery connected → V stays at 200 V. New C = KC = 4 × 10 μF = 40 μF.

U_final = ½ × 40×10⁻⁶ × 200² = 0.8 J = 800 mJ (increases by factor K = 4)

Q3 (MCQ): Two capacitors C₁ = 4 μF and C₂ = 2 μF are connected in series. The ratio of voltage across C₁ to C₂ is:

A) 1:2 B) 2:1 C) 1:1 D) 4:2

Answer: A) 1:2. In series, Q is the same. V = Q/C, so V₁/V₂ = C₂/C₁ = 2/4 = 1:2. The smaller capacitor gets the larger voltage share.

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Capacitors, Capacitance & Dielectrics

1. Capacitance — Definition

2. Parallel Plate Capacitor

Other Capacitor Geometries

3. Combination of Capacitors

Series Combination

Parallel Combination

Worked Example

4. Energy Stored in a Capacitor

5. Effect of Dielectric

Capacitor Quick Reference

Practice Questions

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Practice Questions

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