Physics Class 12 Part I…
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Parallel Plate Combinations Energy

🎓 Class 12 Physics CBSE Theory Ch 2 – Electrostatic Potential and Capacitance ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Parallel Plate Combinations Energy

આ મૂલ્યાંકન આના પર આધારિત હશે: Parallel Plate Combinations Energy

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Parallel Plate Combinations Energy

2.11 The Parallel Plate Capacitor

The most widely used capacitor geometry consists of two flat conducting plates of area \(A\) separated by a distance \(d\) that is small compared with the plate dimensions. One plate carries \(+Q\), the other \(-Q\).

Derivation of C = ε₀A/d

For an infinite sheet of charge density \(\sigma = Q/A\), Gauss's law gives a uniform field \(E = \sigma/\varepsilon_0\) on each side. Between the plates the fields of the two sheets add, while outside they cancel:

\[E_{\text{gap}} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}\quad(\text{uniform, plate-to-plate})\]

The potential difference between the plates is \(V = E\,d\):

\[V = \frac{Qd}{\varepsilon_0 A}\quad\Rightarrow\quad C=\frac{Q}{V}=\boxed{\frac{\varepsilon_0 A}{d}}\]

Capacitance is large when the plates are wide and close together — more area holds more charge; a small gap means a smaller voltage for the same charge.

++++ −−−− E d Area A per plate · C = ε₀A/d
Fig 2.5: A parallel-plate capacitor — the field between the plates is uniform; outside it is essentially zero.

2.12 Effect of a Dielectric

Slip a dielectric slab of constant K completely into the gap. The field inside drops by factor K, therefore so does \(V\), but \(Q\) (if isolated) does not change:

\[C_{\text{with dielectric}} = K\,\frac{\varepsilon_0 A}{d} = K\,C_0\]
MaterialK (approx.)Use
Vacuum / air1.000 / 1.0006Reference
Paper (waxed)3.5Power-factor correction caps
Mica6RF circuits, high-voltage
Glass5 – 10Prototyping, scientific
Ceramic (BaTiO₃)1200+High-density chip caps
Water80Biological capacitance

2.13 Combination of Capacitors

(a) Series Combination

Capacitors are "in series" when they share the same charge \(Q\) but the voltages add:

\[V = V_1 + V_2 + V_3 \;\Rightarrow\; \frac{1}{C_{\text{eq}}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots\]

The equivalent capacitance is smaller than the smallest in the chain.

(b) Parallel Combination

Capacitors are "in parallel" when they share the same voltage \(V\) but the charges add:

\[Q = Q_1 + Q_2 + Q_3 \;\Rightarrow\; C_{\text{eq}} = C_1 + C_2 + C_3 + \ldots\]

The equivalent capacitance is larger than any individual one.

Series C₁ C₂ C₃ 1/Cₑq = 1/C₁+1/C₂+1/C₃ Parallel C₁ C₂ C₃ Cₑq = C₁+C₂+C₃
Fig 2.6: Series (shared Q, voltages add) versus parallel (shared V, charges add).

2.14 Energy Stored in a Capacitor

Charging a capacitor means pushing charge from one plate to the other through an ever-rising potential difference. Suppose, at some instant, the charge is \(q\) and the voltage \(q/C\). Moving an additional \(dq\) requires work \(dW = (q/C)\,dq\). Integrating from 0 to \(Q\):

\[U = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}\]

Using \(Q = CV\), the same energy has three equivalent forms:

\[\boxed{\;U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}\;}\]

Energy Density of the Electric Field

For a parallel-plate capacitor, substitute \(C = \varepsilon_0 A/d\) and \(V = Ed\):

\[U = \tfrac12 CV^2 = \tfrac12\varepsilon_0\frac{A}{d}(Ed)^2 = \tfrac12\varepsilon_0 E^2(Ad)\]

Since \(Ad\) is the volume of the gap, the energy per unit volume is:

\[u = \frac{1}{2}\varepsilon_0 E^2\]

This is a completely general result: electric fields carry energy, at a density proportional to \(E^2\), regardless of how they were produced.

Worked Examples — Plate Capacitor, Combinations, Energy

Example 2.14: NCERT plate capacitor

Plates of area \(6\times 10^{-3}\) m² are separated by a 3 mm air gap. Find its capacitance.

\[C=\frac{\varepsilon_0 A}{d}=\frac{(8.854\times 10^{-12})(6\times 10^{-3})}{3\times 10^{-3}}=\boxed{1.77\times 10^{-11}\,\text{F}\approx 17.7\,\text{pF}}\]

Example 2.15: Three capacitors in series

\(C_1=2\) pF, \(C_2=3\) pF, \(C_3=4\) pF are connected in series across 100 V. Find \(C_{\text{eq}}\), the charge, and the voltage across each.

\[\frac{1}{C_{\text{eq}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\Rightarrow C_{\text{eq}}=\frac{12}{13}\approx 0.923\,\text{pF}\] \[Q=C_{\text{eq}}V = 0.923\times 10^{-12}\times 100 = 9.23\times 10^{-11}\,\text{C}\] \(V_1=Q/C_1=46.15\) V, \(V_2=Q/C_2=30.77\) V, \(V_3=Q/C_3=23.08\) V. (Sum = 100 V ✓)

Example 2.16: Three capacitors in parallel

The same three capacitors (2, 3, 4 pF) are now connected in parallel across 100 V. Find \(C_{\text{eq}}\), total charge, and charge on each.

\(C_{\text{eq}} = 2+3+4 = 9\) pF. Total charge \(Q=9\times 10^{-12}\times 100=9\times 10^{-10}\) C = 0.9 nC. \(Q_1=C_1V=2\times 10^{-10}\) C, \(Q_2=3\times 10^{-10}\) C, \(Q_3=4\times 10^{-10}\) C.

Example 2.17: Mixed (series-parallel) combination

Two \(4\,\mu\)F capacitors are first connected in parallel, and that combination is then put in series with a \(2\,\mu\)F capacitor across a 60 V supply. Find the equivalent capacitance and the voltage across the \(2\,\mu\)F capacitor.

Parallel pair: \(C_{\|}=4+4=8\,\mu\)F. In series with \(2\,\mu\)F: \[\frac{1}{C_{\text{eq}}}=\frac{1}{8}+\frac{1}{2}=\frac{5}{8}\Rightarrow C_{\text{eq}}=1.6\,\mu\text{F}\] Charge on the series combination: \(Q = 1.6\times 10^{-6}\times 60 = 96\,\mu\)C. \(V_{2\mu\text{F}} = Q/2\,\mu\text{F} = 48\) V.

Example 2.18: Energy stored

A 100 μF capacitor is charged to 50 V. How much energy is stored?

\[U=\tfrac12 CV^2 = \tfrac12\times 100\times 10^{-6}\times(50)^2=\boxed{0.125\,\text{J}}\]

Example 2.19: Energy loss when two capacitors are connected

A \(C_1=4\,\mu\)F capacitor is charged to 200 V. It is then connected in parallel to an uncharged \(C_2=4\,\mu\)F capacitor. Find the common voltage, the energy before and after, and the energy lost.

Initial \(Q_i=C_1V_1=8\times 10^{-4}\) C. Initial energy \(U_i=\tfrac12 C_1V_1^2=0.08\) J. Common \(V_f = Q_i/(C_1+C_2)=8\times 10^{-4}/8\times 10^{-6}=100\) V. \(U_f=\tfrac12(C_1+C_2)V_f^2=\tfrac12\times 8\times 10^{-6}\times 100^2=0.04\) J. \[\Delta U = U_i-U_f = \boxed{0.04\,\text{J lost as heat and EM radiation}}\] Half the stored energy is always lost in such charge-sharing, regardless of the capacitance ratio.

Example 2.20: Slab partially filling the gap

A parallel-plate capacitor has plates of area A separated by d. A dielectric slab of thickness \(t < d\) and constant K is inserted parallel to the plates. Find the new capacitance.

The arrangement acts like two capacitors in series: thickness \((d-t)\) in air and thickness \(t\) in dielectric. \[\frac{1}{C}=\frac{d-t}{\varepsilon_0 A}+\frac{t}{K\varepsilon_0 A}\Rightarrow\;\boxed{C=\frac{\varepsilon_0 A}{(d-t)+t/K}}\] When \(t=d\) we recover \(C=K\varepsilon_0 A/d\); when \(K=1\) we recover \(C=\varepsilon_0 A/d\).

Example 2.21: Energy density in field

A parallel-plate capacitor has a field of \(10^{6}\) V/m in its gap of volume \(10^{-4}\) m³. Find the energy stored.

\[u=\tfrac12\varepsilon_0 E^2 = \tfrac12(8.854\times 10^{-12})(10^6)^2 = 4.43\,\text{J/m}^3\] \[U=u\times(\text{volume})=4.43\times 10^{-4}=\boxed{4.43\times 10^{-4}\,\text{J}}\]
Activity — Build a Foil–Paper CapacitorL3 Apply
Predict: You sandwich a sheet of waxed paper between two squares of aluminium foil and connect a multimeter. How does the measured capacitance change as you (a) add more foil-paper layers in parallel, (b) press the sandwich more tightly?
  1. Cut two 10 × 10 cm squares of aluminium foil and one 12 × 12 cm square of waxed paper. Stack them: foil – paper – foil. Clip wires to the two foils.
  2. Read the capacitance on a digital multimeter set to nF.
  3. Build a second identical sandwich and clip it in parallel to the first; read again.
  4. Squeeze the first sandwich with a heavy book and repeat.
Observation: A single sandwich reads ~1–2 nF; two in parallel roughly double the reading; pressing flat reduces \(d\) and bumps C up.

Explanation: \(C = K\varepsilon_0 A/d\). Parallel connection doubles the effective plate area. Pressing reduces \(d\), raising C inversely.

Interactive: Capacitor Combination Solver L3 Apply

Enter up to four capacitances (μF) and choose the connection. The tool returns the equivalent capacitance and total energy at the specified voltage.

Competency-Based Questions

An engineer designs a 12 V flash circuit for a camera. She has 470 μF electrolytic capacitors available. She wants a fast, bright flash (high stored energy) without exceeding each capacitor's 16 V rating.

Q1. L3 Apply Energy stored in a single 470 μF capacitor charged to 12 V is: (2 marks)

\(U=\tfrac12 CV^2=\tfrac12(470\times 10^{-6})(12)^2=\boxed{33.84\,\text{mJ}}\).

Q2. L3 Apply If the engineer connects four of them in parallel across 12 V, what is the total energy available? (2 marks)

\(C_{\text{eq}}=4\times 470=1880\,\mu\)F. \(U=\tfrac12(1880\times 10^{-6})(12)^2=\boxed{0.135\,\text{J}}\) — four times a single cap.

Q3. L1 Remember In a series combination, the quantity that is the same across every capacitor is:

  • A. Voltage
  • B. Charge
  • C. Capacitance
  • D. Energy
B. Charge is common in series; voltage divides in inverse proportion to capacitance.

Q4. L4 Analyse A 5 μF capacitor charged to 200 V is connected in parallel to an uncharged 10 μF capacitor. Find the final voltage and energy lost. (3 marks)

\(Q_i=10^{-3}\) C. \(V_f=10^{-3}/15\times 10^{-6}=66.67\) V. \(U_i=\tfrac12(5\times 10^{-6})(200)^2=0.1\) J; \(U_f=\tfrac12(15\times 10^{-6})(66.67)^2=0.0333\) J. \(\boxed{\Delta U=0.0667\,\text{J lost}}\).

Q5. L3 Apply A parallel-plate capacitor of area 100 cm² and gap 1 mm, air-filled, is charged to 200 V. Compute C, Q and U. (3 marks)

\(C=\varepsilon_0 A/d=(8.854\times 10^{-12})(0.01)/0.001=8.854\times 10^{-11}\) F = 88.5 pF. \(Q=CV=1.77\times 10^{-8}\) C. \(U=\tfrac12 CV^2 = 1.77\times 10^{-6}\) J.

Assertion-Reason Questions

Assertion (A): When a dielectric is inserted in a charged isolated capacitor, the voltage across it drops.

Reason (R): Inserting the dielectric increases C, and with Q fixed, V = Q/C decreases.

  • A. Both A and R true; R explains A.
  • B. Both true; R is not the explanation.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Textbook causal chain: K↑ ⇒ C↑ ⇒ V↓.

Assertion (A): When two capacitors at different potentials are connected by a wire, some energy is always lost.

Reason (R): The transient current dissipates energy as heat and electromagnetic radiation in the connecting wires.

  • A. Both A and R true; R explains A.
  • B. Both true; R is not the explanation.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Charge redistribution is always accompanied by an irreversible energy loss.

Assertion (A): The equivalent capacitance of capacitors in series is smaller than the smallest one.

Reason (R): In series, the effective plate separation adds up, reducing the overall capacitance.

  • A. Both A and R true; R explains A.
  • B. Both true; R is not the explanation.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Series reciprocals add, so the result is below each term.

Did You Know?

Frequently Asked Questions - Parallel Plate Combinations Energy

What is the main concept covered in Parallel Plate Combinations Energy?
In NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance), "Parallel Plate Combinations Energy" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Parallel Plate Combinations Energy useful in real-life applications?
Real-life applications of "Parallel Plate Combinations Energy" from NCERT Class 12 Physics Chapter 2 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Parallel Plate Combinations Energy?
Key formulas in "Parallel Plate Combinations Energy" (NCERT Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 2?
NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance) is structured so each part builds on the previous one. "Parallel Plate Combinations Energy" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Parallel Plate Combinations Energy?
CBSE board questions from "Parallel Plate Combinations Energy" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Parallel Plate Combinations Energy" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part I – NCERT (2025-26)
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