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Most capacitor questions with a dielectric come down to one sentence in the problem: is the battery still connected? The formula C = Kε₀A/d is the same in both cases, but what happens to charge, voltage, field and energy is not. This post follows all four quantities through one capacitor, accounts for every joule, and covers the half-inserted slab that exams like to set.
What does the syllabus say?
The JEE (Main) 2026 syllabus lists "dielectrics and electric polarization" and the "capacitance of a parallel plate capacitor with and without dielectric medium between the plates", plus energy stored in a capacitor. The NEET (UG) 2026 syllabus has the same wording. In the NCERT Class 12 chapter Electrostatic Potential and Capacitance (2026-27 reprint), Section 2.13 explains the effect of a dielectric and Section 2.15 the energy stored. Check the current year's syllabus.
What does a dielectric actually do?
NCERT's Section 2.13 explains it with charge. The field polarises the dielectric, which acts like two sheets of opposite charge on its faces. These bound charges cancel part of the plates' charge, so the field inside becomes E₀/K for the same free charge. The capacitance rises to KC₀. Everything else follows from asking which of Q or V cannot change.
Worked example: one capacitor, both cases
An air-filled parallel plate capacitor has C₀ = 4.0 µF and is charged to V₀ = 50 V. A slab with K = 3 that fills the gap is then slid in.
Before the slab: Q₀ = C₀V₀ = 200 µC and U₀ = ½C₀V₀² = ½ × 4.0 × 10⁻⁶ × 2500 = 5.0 mJ. With the slab, C = 3 × 4.0 = 12 µF.
Case 1: battery disconnected first. Q stays 200 µC.
V = Q/C = 200/12 ≈ 16.7 V, down by a factor of 3.
E = V/d is also down by a factor of 3.
U = Q²/2C = (2.0 × 10⁻⁴)²/(2 × 1.2 × 10⁻⁵) ≈ 1.67 mJ.
Energy lost by the capacitor: 5.0 − 1.67 ≈ 3.33 mJ. The capacitor pulls the slab in, so this energy goes into work on the slab. To insert it slowly, you hold it back and that work flows to you.
Bound charge on each face of the slab: Q(1 − 1/K) = 200 × 2/3 ≈ 133 µC.
Case 2: battery connected. V stays 50 V.
Q = CV = 12 × 50 = 600 µC, three times as much.
E between the plates is still V/d, unchanged. The battery pushes in more free charge until the field is back where it was.
U = ½CV² = ½ × 12 × 10⁻⁶ × 2500 = 15 mJ, an increase of 10 mJ.
Work done by the battery: V × ΔQ = 50 × 400 × 10⁻⁶ = 20 mJ.
The battery supplied 20 mJ but the capacitor kept only 10 mJ. The other 10 mJ is work done on the slab as it is pulled in. For any change at fixed voltage, the battery's work is exactly twice the change in stored energy.
What happens with a slab that only goes halfway in?
If the slab fills the full gap but only half the plate area, the capacitor becomes two capacitors side by side, each with half the area, sharing the same voltage. They are in parallel:
C = C₀/2 + KC₀/2 = C₀(1 + K)/2 = 4.0 × 4/2 = 8.0 µF.
A slab that covers the whole area but only part of the gap is the other case: the layers are in series, as in NCERT's Example 2.8, where a slab of thickness 3d/4 raises C to 4K/(K + 3) times C₀. Ask which way the slab is cut before choosing parallel or series.
Why is this topic so slippery?
Electricity and magnetism concepts are hard for most students even after a course. Maloney and colleagues (2001) gave the 32-question Conceptual Survey of Electricity and Magnetism to more than 5000 introductory students; average posttest scores were only 47% in calculus-based courses and 44% in algebra-based ones. Dielectric problems layer several of those concepts: field, potential, charge and energy. Following each quantity by name, as in the example above, keeps them separate.
Which mistakes should you avoid?
Assuming the field always drops by K. With the battery connected, E between the plates is unchanged.
Assuming energy is conserved inside the capacitor. In both cases energy flows to the slab, and with a battery, from the battery too.
Mixing up bound and free charge. The plates carry free charge; the slab faces carry bound charge.
Using series when the slab is side by side. Split by area means parallel; split by thickness means series.
For networks and charge sharing, see our posts on capacitors and dielectrics problems and charging and sharing capacitors.
A practice routine
Redo the worked example with K = 5, both cases, and fill in Q, V, E and U each time.
Do the energy accounting: capacitor change, battery work and work on the slab.
Remove a slab with the battery connected and decide who does work on whom.
Solve one half-inserted slab by area and one by thickness.
When the picture of bound charges will not form in your head, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Start with step 1, and say which quantity is fixed before you write any number.
