PhysicsElectrostatic potential and capacitanceClass 12

Capacitors, dielectrics and RC circuits: interactive simulation

A capacitor stores charge on two plates separated by an insulator. Its capacitance depends only on its shape: plate area, gap and the material in between.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics C: E&M, A-level Physics · 4 modes in one lab

What you can explore

  1. 1

    Parallel plate

    Plate area, separation and voltage with charge, field and stored energy.
  2. 2

    Dielectric slab

    Slide in a slab of constant K with the battery connected or disconnected.
  3. 3

    RC charging

    Charging and discharging curves, time constant RC and current.
  4. 4

    Series & parallel

    Three capacitors in series or parallel with charge and voltage on each.

Try this

Change the plates and watch the charge and field update, slide in a dielectric and compare what happens with and without the battery, charge it through a resistor, and connect three capacitors in series or parallel.

  1. 1

    Use A = 100 cm², d = 2 mm and 12 V.

    What you should see

    C = ε₀A/d ≈ 44.3 pF, Q ≈ 531 pC, E = 6000 V/m and U ≈ 3.19 nJ.

  2. 2

    Halve the gap.

    What you should see

    C and Q double, and so does the field, because E = V/d.

  3. 3

    In Dielectric slab, insert K = 4 fully with the battery connected.

    What you should see

    C, Q and U all become 4 times larger; V stays 12 V.

  4. 4

    Now disconnect the battery first and insert the slab.

    What you should see

    Q is fixed, so V falls to 3 V and the energy drops to a quarter.

  5. 5

    In RC charging, use 10 V, 10 kΩ and 100 μF.

    What you should see

    τ = 1 s. After 1 s the charge is 632 μC of 1000 μC; the starting current is 1 mA.

  6. 6

    In Series & parallel, connect 2, 3 and 6 μF in series.

    What you should see

    C_eq = 1 μF (smaller than the smallest). In parallel it is 11 μF.

Formulas behind it

Parallel plate capacitor

C=Kε0AdC = \frac{K\varepsilon_0 A}{d}

Charge and energy

Q=CV,U=12CV2=Q22CQ = CV,\qquad U = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}

Slab filling a fraction f of the area

C=C0[1+(K−1)f]C = C_0\left[1 + (K-1)f\right]

RC charging

q(t)=CE(1−e−t/RC)q(t) = CE\left(1 - e^{-t/RC}\right)

Series and parallel

1Cs=∑1Ci,Cp=∑Ci\frac{1}{C_s} = \sum \frac{1}{C_i},\qquad C_p = \sum C_i

Common mistakes

  • Thinking capacitance depends on the charge or the voltage. It depends only on geometry and the dielectric.
  • Mixing up the two dielectric cases: battery connected keeps V constant; disconnected keeps Q constant.
  • Adding capacitors in series like resistors in series. Capacitors in series add as reciprocals.
  • Saying an RC circuit is fully charged after one time constant. It is 63.2 % charged.
Practise 180 questionsCheck your answer instantly, with a worked solution.Electrostatics chapter guideFormula sheet, solved examples and PYQ patterns.Stuck on a capacitors, dielectrics and rc circuits question?Snap it and get a whiteboard video solution.

FAQ

Why does a dielectric increase capacitance?

Its molecules polarise and set up an opposite field inside, reducing the net field and voltage for the same charge, so more charge can be stored per volt.

What is the energy density of an electric field?

u = ½ε₀E² joules per cubic metre (½Kε₀E² inside a dielectric).

Where does half the battery’s energy go when a capacitor charges?

The battery supplies QE but the capacitor stores only ½QE. The other half is dissipated as heat in the resistance, whatever its value.

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