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Exam level6 past questions

Energy stored in capacitors and conservation of charge

Examine how energy is stored, transferred, or lost in capacitors, including during charging, discharging, and redistribution, and the principle of charge conservation.

Why this shows up in the exam

You need to calculate energy changes and understand charge flow in capacitor circuits for NEET.

How NEET tests this

Numerical · 4 QsStatement analysis · 2 QsMulti-concept

Learn the idea

Energy stored in a capacitor equals the work needed to bring charge from zero to its final value; for a parallel‑plate capacitor it reduces to U = ½ ε₀ E² A d, the field energy filling the volume between the plates.

🧠 Memory hook: Half‑ε₀E² fills the capacitor’s volume – think of the energy as ‘half the vacuum field energy’ occupying the space A·d.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • C = ε₀A/d (vacuum parallel‑plate capacitor)
  • V = E d (field times separation)
  • U = ½ C V² = ½ Q²/C = ½ ε₀ E² A d
  • Q = C V
  • Charge is conserved in an isolated system – total charge before and after connecting capacitors remains the same

How to approach it

  1. 1Identify what is given – usually E, d and A or C and V
  2. 2Write C = ε₀A/d and V = E d if needed
  3. 3Use U = ½ C V² (or directly U = ½ ε₀ E² A d) to get the energy
  4. 4If capacitors are connected, apply charge conservation: ΣQ_initial = ΣQ_final before using the energy formula

Worked example — watch it click

A parallel plate condenser has a uniform electric field E(V/m) in the space between the plates. If the distance between the plates is d(m) and the area of each plate is A(m²), the energy (joules) stored in the condenser is:

  • A)E²Ad/ε₀
  • B)(1/2)ε₀E²
  • C)ε₀EAd
  • ✅(1/2)ε₀E²Ad

The concept behind this problem

The question forces you to translate the given uniform field into voltage and capacitance, then apply the universal energy formula, revealing the compact expression ½ ε₀E²Ad.

Step by step

  1. 1This is identical to Q29.
  2. 2For a parallel plate capacitor with field E, separation d, and area A: C = ε₀A/d, V = Ed.
  3. 3Energy U = (1/2)CV² = (1/2)(ε₀A/d)(Ed)² = (1/2)ε₀E²Ad.

Watch out

Students often drop the ½ and pick ε₀E²Ad as the answer.

Common slip-ups that cost marks

  • •Missing the factor ½ gives twice the correct energy
  • •Using ε (dielectric constant) instead of ε₀ for a vacuum capacitor
  • •Confusing E (V m⁻¹) with V (volts) and forgetting to multiply by d

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 6NEET 2012

A parallel plate condenser has a uniform electric field E(V/m) in the space between the plates. If the distance between the plates is d(m) and the area of each plate is A(m²), the energy (joules) stored in the condenser is:

Push further

More challenging

8 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 8

A 500 pF capacitor is charged to 100 V. It is then disconnected from the battery and connected in parallel to an uncharged 500 pF capacitor. A student calculates the final energy of the system by using the initial voltage and the total capacitance. What mistake did the student likely make?

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