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Capacitor flash cards

Master Capacitor through 89 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Capacitor, question and answer

24 of this chapter's 89 cards, laid out open so you can read straight through. The remaining 65 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.Define capacitance of a conductor/capacitor.

    Capacitance C=Q/VC = Q/V is the charge stored per unit potential difference. It depends only on geometry and the surrounding medium, not on QQ or VV.

    Hint: Ratio of charge to potential.

  2. 2.What is the SI unit of capacitance and its definition?

    The farad (F): 1F=1C/V1\,\text{F} = 1\,\text{C/V}. One farad is a very large unit; practical values are in μF\mu\text{F}, nFn\text{F} and pFp\text{F}.

    Hint: Coulomb per volt.

  3. 3.Give the capacitance of an isolated spherical conductor of radius RR.

    C=4πε0RC = 4\pi\varepsilon_0 R. It behaves like a sphere-and-infinity capacitor; capacitance grows linearly with radius.

    Hint: Use V=Q4πε0RV = \frac{Q}{4\pi\varepsilon_0 R}.

  4. 4.State the parallel plate capacitor formula (vacuum) and name each symbol.

    C=ε0AdC = \dfrac{\varepsilon_0 A}{d}, where AA is plate area, dd the separation, ε0\varepsilon_0 the permittivity of free space.

    Hint: Area over gap.

  5. 5.How does the capacitance of a parallel plate capacitor change if plate separation dd is doubled (area fixed)?

    C1/dC \propto 1/d, so doubling dd halves CC.

    Hint: Inverse dependence on dd.

  6. 6.What field exists between the plates of a parallel plate capacitor and its value?

    A nearly uniform field E=σε0=Qε0A=VdE = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A} = \dfrac{V}{d}, directed from + to − plate.

    Hint: σ/ε0\sigma/\varepsilon_0.

  7. 7.Why must the two plates of a capacitor carry equal and opposite charges +Q+Q and Q-Q?

    Charge +Q+Q on one plate induces Q-Q on the facing plate. The field is confined between them, and this configuration is what stores energy; the 'charge on capacitor' means the magnitude QQ.

    Hint: Induction between facing plates.

  8. 8.Define dielectric constant (relative permittivity) KK.

    K=ε/ε0=Cmedium/CvacuumK = \varepsilon/\varepsilon_0 = C_{\text{medium}}/C_{\text{vacuum}}. It is the factor by which capacitance increases when the gap is filled with the dielectric (K1K \geq 1).

    Hint: Ratio of permittivities.

  9. 9.A parallel plate capacitor is completely filled with a dielectric of constant KK. New capacitance?

    C=Kε0Ad=KC0C = \dfrac{K\varepsilon_0 A}{d} = K C_0. Capacitance increases KK times.

    Hint: Multiply vacuum value by KK.

  10. 10.Why does inserting a dielectric increase capacitance?

    The dielectric polarizes, producing bound charges that partially cancel the field, reducing EE (and VV) for the same QQ. Since C=Q/VC = Q/V, lower VV means higher CC.

    Hint: Polarization reduces internal field.

  11. 11.What is dielectric strength and why does it matter for capacitors?

    Dielectric strength is the maximum field a material can withstand before breakdown (conduction). It sets the maximum safe voltage Vmax=EbreakdowndV_{\max} = E_{\text{breakdown}}\,d a capacitor can hold.

    Hint: Breakdown field limit.

  12. 12.Derive capacitance of a parallel plate capacitor with a dielectric slab of thickness tt (t<dt<d) and constant KK.

    C=ε0Adt+t/KC = \dfrac{\varepsilon_0 A}{d - t + t/K}. The slab acts like reducing the effective air gap by t(11/K)t(1 - 1/K).

    Hint: Series of air and dielectric regions.

  13. 13.A conducting slab of thickness tt is inserted (parallel to plates, t<dt<d). New capacitance?

    C=ε0AdtC = \dfrac{\varepsilon_0 A}{d - t}. A conductor is the KK\to\infty limit, so it just removes thickness tt from the gap.

    Hint: Set KK\to\infty in the slab formula.

  14. 14.For a partial dielectric slab, does the answer depend on where (which position) the slab sits in the gap?

    No. For a slab parallel to the plates, only its thickness tt matters, not its distance from either plate (as long as it stays fully inside).

    Hint: Series capacitors add reciprocally regardless of order.

  15. 15.Two capacitors C1,C2C_1, C_2 in series: equivalent capacitance and charge relation?

    1Ceq=1C1+1C2\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}. Same charge QQ on each; voltages add: V=V1+V2V = V_1 + V_2.

    Hint: Reciprocals add; equal charge.

  16. 16.Two capacitors C1,C2C_1, C_2 in parallel: equivalent capacitance and voltage relation?

    Ceq=C1+C2C_{eq} = C_1 + C_2. Same voltage VV across each; charges add: Q=Q1+Q2Q = Q_1 + Q_2.

    Hint: Capacitances add; equal voltage.

  17. 17.For series capacitors, across which capacitor is the voltage largest?

    Across the smallest capacitor. Since QQ is common, V=Q/CV = Q/C, so smaller CC takes larger VV.

    Hint: V1/CV \propto 1/C at fixed QQ.

  18. 18.For two capacitors in series, give CeqC_{eq} as a single fraction.

    Ceq=C1C2C1+C2C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} (product over sum).

    Hint: Analogous to parallel resistors.

  19. 19.Energy stored in a charged capacitor — give the three equivalent forms.

    U=12CV2=Q22C=12QVU = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} = \dfrac{1}{2}QV.

    Hint: Use Q=CVQ = CV to switch forms.

  20. 20.Why is the energy stored 12QV\frac{1}{2}QV and not QVQV?

    Charge is transferred gradually; the potential rises from 00 to VV as charging proceeds, so the average work per unit charge is V/2V/2. Integrating Vdq=Q2/2C\int V\,dq = Q^2/2C.

    Hint: Average potential during charging.

  21. 21.Define energy density of the electric field between capacitor plates.

    u=12ε0E2u = \dfrac{1}{2}\varepsilon_0 E^2 (vacuum), or 12Kε0E2\dfrac{1}{2}K\varepsilon_0 E^2 in a dielectric. It is energy per unit volume of the field.

    Hint: Half epsilon E squared.

  22. 22.Capacitance of a spherical capacitor (inner radius aa, outer bb, vacuum).

    C=4πε0abbaC = 4\pi\varepsilon_0\,\dfrac{ab}{b - a}.

    Hint: Integrate field between shells for VV.

  23. 23.In the spherical capacitor formula, what limit recovers the isolated sphere result?

    Let bb \to \infty: C=4πε0abba4πε0aC = 4\pi\varepsilon_0\,\dfrac{ab}{b-a} \to 4\pi\varepsilon_0 a, the isolated sphere of radius aa.

    Hint: Push the outer shell to infinity.

  24. 24.Capacitance of a cylindrical capacitor (length LL, radii aa inner and bb outer).

    C=2πε0Lln(b/a)C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}, assuming LbL \gg b (edge effects ignored).

    Hint: Field of a line charge, integrate for VV.

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