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Capacitor problems reward one habit above all: deciding what stays constant before you calculate. Insert a dielectric with the battery connected, and the charge goes up. Insert it after disconnecting, and the voltage goes down. The formulas are the same, the answers opposite. This post works through networks, dielectrics and energy with numbers, and shows the checks that catch errors.
What is in the syllabus?
The JEE (Main) 2026 syllabus lists dielectrics and electric polarisation, capacitors in series and parallel, the parallel plate capacitor with and without a dielectric, and energy stored in a capacitor. The JEE (Advanced) 2026 brochure lists the same ideas. The CBSE 2026-27 curriculum includes them too, with energy stored marked "no derivation, formulae only". Check the current documents for your exam.
What sets the capacitance?
The NCERT Class 12 chapter Electrostatic Potential and Capacitance derives C = ε₀A/d for two large parallel plates of area A a distance d apart. It also notes that a farad is a very big unit, so you will usually see µF, nF and pF. The constant ε₀, the vacuum permittivity, has the CODATA value from NIST of 8.854 × 10⁻¹² F/m.
Example 1. Plates of area 0.020 m² (200 cm²) are 1.0 mm apart. C = 8.854 × 10⁻¹² × 0.020 / 1.0 × 10⁻³ ≈ 1.77 × 10⁻¹⁰ F, or about 177 pF. Filled with a dielectric of K = 5, it becomes about 885 pF.
NCERT defines the dielectric constant as the factor, greater than 1, by which capacitance increases from its vacuum value when the dielectric fills the space between the plates. The dielectric's own charges partly cancel the field inside, so the same charge gives a smaller voltage.
How do you solve a capacitor network?
Parallel capacitors share the same voltage, so their capacitances add. Series capacitors carry the same charge, so 1/C adds. That is the opposite of resistors, which is the most common slip.
Example 2. A 2 µF and a 3 µF capacitor are in series. That pair is in parallel with a 1.8 µF capacitor, and the whole network is connected across 10 V.
Series pair: 1/C = 1/2 + 1/3 = 5/6, so C = 1.2 µF.
Total: 1.2 + 1.8 = 3.0 µF.
Charge on the series pair: 1.2 × 10 = 12 µC, the same on both. So the 2 µF has 6 V and the 3 µF has 4 V. They add to 10 V, as they should.
Charge on the 1.8 µF: 18 µC. Total: 12 + 18 = 30 µC = 3.0 µF × 10 V. It checks.
Energy stored: ½ × 3.0 × 10⁻⁶ × 10² = 1.5 × 10⁻⁴ J, or 150 µJ.
What happens when you insert a dielectric?
Example 3. A 5 µF capacitor is charged to 20 V, so Q = 100 µC and U = ½ × 5 × 10⁻⁶ × 20² = 1.0 mJ. A dielectric of K = 4 is then slid in to fill the gap. C becomes 20 µF.
Battery disconnected first. Q stays 100 µC. V = Q/C = 5 V. U = Q²/2C = (10⁻⁴)² / (2 × 2 × 10⁻⁵) = 0.25 mJ. Energy falls by a factor K; the capacitor pulls the slab in and does work on it.
Battery still connected. V stays 20 V. Q = 20 × 20 = 400 µC. U = ½ × 20 × 10⁻⁶ × 20² = 4.0 mJ. Energy rises by a factor K, supplied by the battery.
A slab that does not fill the gap. If a slab of thickness t sits between plates d apart, C = ε₀A / (d − t + t/K). With d = 4 mm, t = 2 mm and K = 4, the denominator is 2 + 0.5 = 2.5 mm, so C is 4/2.5 = 1.6 times its vacuum value. NCERT's Example 2.8 does the same calculation for a slab three-quarters of the gap.
Where does energy go when capacitors share charge?
Example 4. A 2 µF capacitor charged to 10 V (100 µJ) is connected to an identical uncharged one. Charge spreads equally: 20 µC becomes 10 µC each, so the common voltage is 5 V. The energy is now ½ × 4 × 10⁻⁶ × 5² = 50 µJ. Half the energy is lost to heat in the wires and to radiation, however small their resistance. Charge is conserved; energy stored is not.
Why do these ideas feel slippery?
Students often learn the formulas without a picture of charges moving. Guisasola, Zubimendi and Zuza (2010) designed a teaching sequence on capacitance for first-year university students, built around common student conceptions. Compared with a control group, it helped students connect the movement of charges with the energy analysis of the system. When you solve a problem, say in words where charge goes and where energy comes from, not just which formula to use.
Which mistakes cost marks?
Swapping series and parallel rules because they are the reverse of resistors.
Forgetting the battery question in dielectric problems.
Prefixes. µF is 10⁻⁶ F and pF is 10⁻¹² F.
Assuming energy is conserved when capacitors are joined. It is not.
To picture the field between the plates, see our guide to visualising electric fields. Next, see how capacitors charge over time in our guide to RC circuits and the time constant. If a step in a network solution does not make sense, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Day 1: Five networks, checking that charges and voltages add correctly.
Day 2: Four dielectric problems, two with the battery connected and two without.
Day 3: Three partial-slab problems and two on charge sharing.
Day 4: A timed mixed set, with the "battery connected?" line written first.
Tonight, redo Example 3 with K = 2 and check that the two energies come out as 0.5 mJ and 2.0 mJ.
