In a parallel plate capacitor with air between the plates, each plate has an area of 6×10^-3 m^2 and the separation between the plates is 2 mm. a) Calculate the capacitance of the capacitor, b) If this capacitor is connected to 100 V supply, what would be the charge on each plate

Question

In a parallel plate capacitor with air between the plates, each plate has an area of 6×10^-3 m^2 and the separation between the plates is 2 mm. a) Calculate the capacitance of the capacitor, b) If this capacitor is connected to 100 V supply, what would be the charge on each plate? c) How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?

Accepted Answer

a) $C=\dfrac{\epsilon_0A}{d}=\dfrac{8.85\times10^{-12}\times6\times10^{-3}}{2\times10^{-3}}=8.85\times10^{-12}\times3=2.655\times10^{-11}$ F. b) At $V=100$ V: $Q=CV=2.655\times10^{-11}\times100=2.655\times10^{-9}$ C. c) A 2 mm mica sheet (k=6) inserted fully fills the 2 mm gap ( $t=d$ ), so by special case (1) of section 8.10.2, $C'=kC=6\times2.655\times10^{-11}=1.593\times10^{-10}$ F. Since the voltage SUPPLY REMAINS CONNECTED (V is held fixed at 100 V, not the charge), the new charge is $Q'=C'V=1.593\times10^{-10}\times100=1.593\t …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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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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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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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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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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CBSE 2026Set ANNUAL1 mark
$\text{m}^2$
›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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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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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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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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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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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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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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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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