CUET UG Physics Electrostatics Capacitors Capacitance Combinations Energy Stored Dielectrics

Must-Know (15–20 detailed bullets)

• Capacitance of a parallel plate capacitor is given by $ C = \frac{\varepsilon_0 A}{d} $, where $ A $ is plate area and $ d $ is separation; for example, if $ A = 0.5 \, \text{m}^2 $, $ d = 0.01 \, \text{m} $, $ C \approx 4.43 \times 10^{-10} \, \text{F} $.
• When a dielectric of dielectric constant $ K $ is inserted fully between plates, capacitance becomes $ C' = K C_0 $; e.g., with mica ($ K = 6 $), capacitance increases sixfold.
• Energy stored in a capacitor is $ U = \frac{1}{2} CV^2 = \frac{Q^2}{2C} = \frac{1}{2} QV $; a 10 µF capacitor charged to 100 V stores $ U = 0.05 \, \text{J} $.
• In series combination, reciprocal of equivalent capacitance is sum of reciprocals: $ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots $; two 4 µF capacitors in series give $ C_{eq} = 2 \, \mu\text{F} $.
• In parallel combination, $ C_{eq} = C_1 + C_2 + \cdots $; three 2 µF capacitors in parallel yield $ C_{eq} = 6 \, \mu\text{F} $.
• Charge remains constant when a charged capacitor is disconnected before inserting a dielectric; voltage decreases by factor $ K $.
• Voltage remains constant when battery remains connected during dielectric insertion; charge increases by factor $ K $.
• Polarization of dielectrics reduces the effective electric field inside: $ E = \frac{E_0}{K} $, where $ E_0 $ is field without dielectric.
• Dielectric constant $ K $ is always greater than 1; for vacuum, $ K = 1 $; for water, $ K \approx 80 $ (verify from NCERT).
• Electric field between plates of a parallel plate capacitor is $ E = \frac{\sigma}{\varepsilon_0} $ without dielectric; with dielectric, $ E = \frac{\sigma}{K \varepsilon_0} $.
• Capacitance depends only on geometry and dielectric medium, not on charge or voltage.
• Energy density in electric field is $ u = \frac{1}{2} \varepsilon_0 E^2 $; with dielectric, it becomes $ u = \frac{1}{2} K \varepsilon_0 E^2 $.
• For a spherical capacitor with inner radius $ a $, outer radius $ b $: $ C = 4\pi\varepsilon_0 \frac{ab}{b-a} $.
• For an isolated spherical conductor of radius $ R $, $ C = 4\pi\varepsilon_0 R $; Earth ($ R \approx 6.4 \times 10^6 \, \text{m} $) has $ C \approx 711 \, \mu\text{F} $.
• In series combination, charge on each capacitor is same; in parallel, voltage across each is same.
• When two capacitors are connected in parallel after charging separately, common potential is $ V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} $.
• If a metal slab (thickness $ t $) is inserted between plates, capacitance becomes $ C = \frac{\varepsilon_0 A}{d - t} $; acts like reduced separation.
• Work done by battery in charging a capacitor is $ QV $, but energy stored is only $ \frac{1}{2} QV $; rest is lost as heat.
• Van de Graaff generator uses the principle that charge resides on outer surface of hollow conductor — related to capacitance of spherical shells.
• Dielectrics can be polar or non-polar; polar dielectrics (e.g., water) have permanent dipole moments.

Difficulty Level

Intermediate — combines conceptual understanding of electric fields, charge conservation, and algebraic manipulation in combinations.

Common CUET Traps

• Trap: Assuming energy is conserved when two capacitors are connected.
  Avoid: Energy is lost as heat; use charge conservation, not energy conservation.

• Trap: Thinking dielectric constant can be less than 1.
  Avoid: $ K \geq 1 $ always; $ K = 1 $ for vacuum, greater for all materials.

• Trap: Believing voltage divides equally in series capacitors regardless of capacitance.
  Avoid: Voltage is inversely proportional to capacitance in series: $ V \propto \frac{1}{C} $.

Practice MCQs

Q1. A parallel plate capacitor has capacitance 10 µF with air between plates. A dielectric slab of constant $ K = 5 $ is inserted while the battery remains connected. The new capacitance is:
A. 2 µF
B. 10 µF
C. 50 µF
D. 0.5 µF
Answer: C
Explanation: With battery connected, $ C' = K C_0 = 5 \times 10 = 50 \, \mu\text{F} $.
Why others fail: Option A assumes inverse relationship, common confusion when battery is disconnected.

Q2. Three capacitors each of 3 µF are connected in series. The equivalent capacitance is:
A. 9 µF
B. 3 µF
C. 1 µF
D. 6 µF
Answer: C
Explanation: $ \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 $, so $ C_{eq} = 1 \, \mu\text{F} $.
Why others fail: Option A adds them directly, confusing series with parallel.

Q3. A 4 µF capacitor is charged to 200 V. The energy stored in it is:
A. 0.08 J
B. 0.16 J
C. 0.32 J
D. 0.04 J
Answer: A
Explanation: $ U = \frac{1}{2} C V^2 = \frac{1}{2} \times 4 \times 10^{-6} \times (200)^2 = 0.08 \, \text{J} $.
Why others fail: Option B doubles the correct value, possibly due to missing 1/2 factor.

Q4. Two capacitors $ C_1 = 2 \, \mu\text{F} $ and $ C_2 = 4 \, \mu\text{F} $ are charged to 100 V and 50 V respectively. They are then connected positive to positive. The common potential is:
A. 66.7 V
B. 75 V
C. 50 V
D. 100 V
Answer: A
Explanation: $ V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} = \frac{(200 + 200)}{6} = 66.7 \, \text{V} $.
Why others fail: Option B averages voltages directly, ignoring capacitance weights.

Q5. A parallel plate capacitor is charged and then disconnected from the battery. A dielectric slab of $ K = 4 $ is inserted fully between plates. Which of the following remains unchanged?
A. Electric field
B. Energy stored
C. Charge
D. Capacitance
Answer: C
Explanation: When disconnected, charge remains conserved.
Why others fail: Option D is tempting because capacitance changes, but students may misremember what stays constant.

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ $ C = \frac{Q}{V} $ — definition of capacitance.
• ⚠️ In series: same charge, different voltage; in parallel: same voltage, different charge.
• ⚠️ Energy stored: $ \frac{1}{2} CV^2 $ — most frequently tested formula.
• ⚠️ Dielectric increases capacitance by factor $ K $, reduces electric field by $ K $.
• ⚠️ With battery connected: $ V $ constant; without battery: $ Q $ constant.
• ⚠️ $ C_{\text{sphere}} = 4\pi\varepsilon_0 R $ — for isolated conducting sphere.
• ⚠️ For series capacitors: smallest capacitor has largest voltage drop.
• ⚠️ Work done by battery = $ QV $; energy stored = $ \frac{1}{2} QV $.
• ⚠️ Polar dielectrics have permanent dipoles; non-polar develop induced dipoles.
• ⚠️ $ E = \frac{\sigma}{\varepsilon_0} $ — field due to infinite sheet (also applies to parallel plates).
• ⚠️ When dielectric fills space, $ C = K \frac{\varepsilon_0 A}{d} $.
• ⚠️ If metal slab inserted, $ C = \frac{\varepsilon_0 A}{d - t} $ — effective decrease in plate separation.
• ⚠️ Energy density: $ \frac{1}{2} \varepsilon_0 E^2 $ — same form as in electromagnetic waves.
• ⚠️ No current flows through capacitor in steady DC — open circuit.
• ⚠️ Common potential: $ V = \frac{\text{Total charge}}{\text{Total capacitance}} $.
• ⚠️ Dielectric constant of vacuum = 1; air ≈ 1.0006 — often approximated as 1.
• ⚠️ Capacitance independent of material of plates — depends only on shape, size, medium.
• ⚠️ $ K = \frac{C}{C_0} $ — operational definition of dielectric constant.
• ⚠️ Mnemonic: “Battery ON → V constant; Battery OFF → Q constant”.
• ⚠️ Van de Graaff generator uses high capacitance of large sphere to store charge.