Q.Explain the concept of a parallel plate capacitor. State its any 'two' applications.

Showing the 12 most recent of 19 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.A capacitor of capacitance 1 μF is kept at 1 V potential difference between the plates. The charge on the charged capacitor will be (A) 1 C (B) zero (C) 1 μC (D) none of these

   ›Reveal solutionSolution

   Q = CV = 1 μF × 1 V = 1 μC.

   The charge on a capacitor is $Q = CV$.

   Here $C = 1\ \mu\text{F} = 1\times10^{-6}$ F and $V = 1$ V, so …

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2. CBSE 2026Set A1 markMCQ
   Q.If the area of plates of a capacitor is halved, then the capacitance will (A) become double (B) become half (C) remain same (D) become four times

   ›Reveal solutionSolution

   Capacitance is proportional to plate area (C = ε₀A/d), so halving the area halves C.

   The capacitance of a parallel-plate capacitor is

   $C = \frac{\varepsilon_0 A}{d}$ …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The unit of electric capacitance is

   (a) volt

   (b) newton

   (c) farad

   (d) ampere

   ›Reveal solutionSolution

   Capacitance is defined as charge stored per unit potential difference, and its SI unit is the farad.

   Capacitance of a conductor (or a capacitor) is defined by C = Q / V, where Q is the charge stored and V is the potential difference across it. Since Q is measured in coulombs and V in volts, the unit of capacitance is coulomb/volt, which is given the special name farad (F), after Michael F …

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4. CBSE 2026Set ANNUAL1 mark
   Q.Write the SI unit of capacitance of capacitor.

   ›Reveal solutionSolution

   Capacitance $C=Q/V$, so its SI unit is coulomb per volt, named the farad (F).

   Capacitance is defined as the charge stored per unit potential difference, $C=\dfrac{Q}{V}$. Since charge is measured in coulombs (C) and potential difference in volts (V), the SI unit of capacitance is coulomb/volt, which is given the special name farad, symbol F, in honour of Michael Faraday. One farad is a very larg …

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5. CBSE 2026Set ANNUAL1 mark
   Q.A parallel-plate capacitor of plate area 2 $\text{m}^2$, separated by a distance of 1 mm is immersed in castor oil of dielectric constant 4.7. What is its capacitance? OR Two capacitors each charged with $4.8\times10^{-8}$ C are connected in parallel to each other. If the potential difference of one of the capacitors is 12 V, calculate the total energy stored in both the capacitors.

   ›Reveal solutionSolution

   A dielectric-filled parallel-plate capacitor's capacitance is $C=K\varepsilon_0A/d$; substituting the given area, separation and dielectric constant of castor oil gives $C\approx8.32\times10^{-8}$ F.

   Formula for a capacitor filled with a dielectric

   For a parallel-plate capacitor completely filled with a dielectric of dielectric constant $K$, plate area $A$, and plate separation $d$:

   $C=\frac{K\varepsilon_0 A}{d}$

   Substituting the given values

   $A=2\text{ m}^2$, $d=1\text{ mm}=1\times10^{-3}\text{ m}$, $K=4.7$ (castor oil), $\varepsilon_0=8.85\times10^{-12}\text{ F/m}$:

   $C=\frac{4.7\times(8.85\times10^{-12})\times2}{1\times10^{-3}}$

   $C=\frac{4.7\times2\times8.85\times10^{-12}}{10^{-3}}=4.7\times2\times8.85\times10^{-9}$

   $C \approx 83.2\times10^{-9}\text{ F} = 8.32\times10^{-8}\text{ F}$

   …

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6. CBSE 2025Set D1 markMCQ
   Q.Which one of the following is unit of capacity? (A) coulomb (B) ampere (C) volt (D) coulomb/volt

   ›Reveal solutionSolution

   Capacitance is charge stored per unit potential, C = Q/V, so its SI unit is coulomb/volt (the farad).

   The capacity (capacitance) of a conductor or capacitor is defined as the charge stored per unit rise in potential:

   $C = \frac{Q}{V}$

   Therefore its unit is (unit of charge)/(unit of potential) = coulomb/volt. One coulomb/volt is given the name farad (F).

   …

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7. CBSE 2025Set D1 markMCQ
   Q.The capacity of any condenser does not depend upon (A) shape of plates (B) size of plates (C) charge on plates (D) distance between plates

   ›Reveal solutionSolution

   Capacitance depends only on the geometry (plate size, shape, separation) and the dielectric — not on how much charge is placed on it.

   For a parallel-plate capacitor,

   $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$

   where A is the plate area, d the separation and $\varepsilon_r$ the dielectric constant. Capacitance therefore depends on the shape and size of the plates, the distance between them, and the medium — all geometric/material factors …

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8. CBSE 2024Set A1 markMCQ
   Q.Picofarad is the unit of (A) electric charge (B) intensity of electric field (C) electric capacity (D) electric flux

   ›Reveal solutionSolution

   Picofarad = 10⁻¹² farad, and the farad is the unit of capacitance.

   Capacitance is measured in farads (F). Common sub-multiples are microfarad (μF = 10⁻⁶ F), nanofarad (nF = 10⁻⁹ F) and picofarad (pF = 10⁻¹² F).

   …

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9. CBSE 2024Set A1 markMCQ
   Q.Capacity of any condenser does not depend upon (A) shape of plates (B) size of plates (C) charges on plates (D) distance between plates

   ›Reveal solutionSolution

   C depends on plate area, separation, shape and dielectric — not on the charge on the plates.

   For a parallel-plate capacitor $C = \dfrac{\varepsilon_0 \varepsilon_r A}{d}$. Capacitance is fixed by the geometry (plate size/shape, separation d) and the dielectric.

   …

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10. CBSE 2024Set A1 markMCQ
    Q.Which of the following is blocked by a capacitor? (A) AC (B) DC (C) Both AC and DC (D) Neither AC nor DC

    ›Reveal solutionSolution

    A charged capacitor stops steady (DC) current but passes AC via continuous charge/discharge.

    The capacitive reactance is $X_C = \dfrac{1}{2\pi f C}$.

    …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The capacitance of a parallel plate capacitor depends upon

    (a) thickness of the plate

    (b) mass of the plate

    (c) density of the plate

    (d) area of the plate

    ›Reveal solutionSolution

    The capacitance of a parallel plate capacitor is a purely geometric/electrical quantity, C = epsilon0 K A / d, so among the choices only the plate area matters.

    For a parallel plate capacitor with plate area A, plate separation d, and a dielectric of constant K filling the gap,

    $C = \dfrac{K\epsilon_0 A}{d}$

    …

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12. CBSE 2024Set ANNUAL1 mark
    Q.Define the S.I. unit of capacitance.

    ›Reveal solutionSolution

    The SI unit of capacitance is the farad, defined via C = Q/V.

    Capacitance is defined as $C = Q/V$, the charge stored per unit potential difference. The SI unit is the farad (F), named after Michael Faraday:

    $1\ \text{farad} = \dfrac{1\ \text{coulomb}}{1\ \text{volt}}$

    …

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