Q.A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are (A) Constant, decreases, decreases (B) Increases, decreases, decreases (C) Constant, decreases, increases (D) Constant, increases, decreases

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

   $Q = (1\times10^{-6}\ \text{F})(1\ \text{V}) = 1\times10^{-6}\ \text{C} = 1\ \mu\text{C}.$

   ✓Final answer

   (C) 1 μC.

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

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

   with $A$ the plate area and $d$ the separation. Capacitance is directly proportional to $A$, so if the area is reduced to $A/2$ while $d$ stays the same, the capacitance becomes

   $C' = \frac{\varepsilon_0 (A/2)}{d} = \frac{C}{2}.$

   ✓Final answer

   (B) become half.

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 Faraday. (Volt is the unit of potential, newton is the unit of force, and ampere is the unit of current - none of these is the unit of capacitance.)

   ✓Final answer

   (c) farad.

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 large capacitance, so practical capacitors are usually rated in microfarad ($\mu\text{F}=10^{-6}\ \text{F}$) or picofarad ($\text{pF}=10^{-12}\ \text{F}$).

   ✓Final answer

   Farad (F)

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), $\varepsilon_0=8.85\times10^{-12}\text{ F/m}$:

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

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

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

   ✓Final answer

   $C\approx8.32\times10^{-8}\text{ F}\approx0.083\,\mu\text{F}$ (about $83.2$ nF).

   ---

   Alternative (Or):

   Since both capacitors carry the same charge and the connection is a parallel one between capacitors already at the same potential, the total energy is simply the sum of the two individual $\frac12QV$ energies, giving $5.76\times10^{-7}$ J.

   Reading the data

   Both capacitors are charged with the same charge $Q=4.8\times10^{-8}$ C. We are told the potential difference of one of them is $V=12$ V. Since both capacitors carry the identical charge $Q$ and no further information distinguishes them, they must be identical capacitors (same capacitance), so both are at the same potential difference, $V=12$ V (a capacitor's voltage is fixed once its charge and capacitance are fixed: $V=Q/C$). Connecting two capacitors that already sit at the same potential in parallel causes no redistribution of charge (no current flows, since there is no potential difference between them to drive one) — so the total stored energy is simply the sum of what each capacitor already stored individually.

   Energy stored in each capacitor

   $U=\frac12 QV$

   $U=\frac12\times(4.8\times10^{-8})\times12 = \frac12\times5.76\times10^{-7}=2.88\times10^{-7}\text{ J}$

   Total energy for both capacitors

   $U_{\text{total}}=2\times2.88\times10^{-7}=5.76\times10^{-7}\text{ J}$

   ✓Final answer

   Total energy stored in both capacitors $\approx 5.76\times10^{-7}\text{ J}$.

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).

   Checking the wrong options: coulomb is the unit of charge, ampere the unit of current, and volt the unit of potential — none of these is capacitance.

   ✓Final answer

   (D) coulomb/volt.

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.

   Although Q and V both change when you charge the capacitor, their ratio C = Q/V stays fixed. So the capacitance does not depend on the charge on the plates.

   ✓Final answer

   (C) charge on plates.

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).

   Hence picofarad is a unit of electric capacity (capacitance).

   ✓Final answer

   (C) electric capacity.

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.

   Although Q = CV, increasing the charge increases the voltage proportionally so that C = Q/V stays constant. Thus capacity does NOT depend on the charge on the plates.

   ✓Final answer

   (C) charges on plates.

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 $X_C = \dfrac{1}{2\pi f C}$.

    For DC the frequency f = 0, so X_C → ∞: once charged, no steady current flows — DC is blocked. For AC, f ≠ 0 gives a finite reactance, and the plates charge and discharge each cycle, so AC effectively passes through.

    ✓Final answer

    (B) DC.

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,

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

    This shows C depends on the area of overlap of the plates (A), their separation (d), and the dielectric medium between them (K). It does not depend on the thickness, mass, or density of the plates themselves - those are mechanical properties of the conducting plate, not the field/geometry between them. Among the given options only 'area of the plate' is a genuine determinant of C.

    ✓Final answer

    (d) area of the plate.

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:

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

    i.e. a capacitor has a capacitance of one farad if a charge of one coulomb raises its potential by one volt. (Since the farad is a very large practical unit, capacitances are usually expressed in microfarad ($10^{-6}$ F) or picofarad ($10^{-12}$ F).)

    ✓Final answer

    The farad (F): 1 F = 1 C/1 V.