Case study (Capacitors): A capacitor is a system of two conductors separated by an insulator, with charges Q a

Question

Figure — Figure — 55/6/1 Q29 — FigureFigure — 55/6/1 Q29

Q

(i) A capacitor of capacitance C, plate area A and separation d, is filled with air [Fig.(a)]. The separation is increased to 2d and one plate is shifted as shown in Fig.(b). The capacitance of the new system is: (A)

\dfrac{C}{4}

(B)

\dfrac{C}{2}

(C)

2C

(D)

4C

(ii) A slab (area A, thickness

d_1

) of a linear dielectric of dielectric constant K is inserted between charged plates (charge density

\sigma

) of a parallel plate capacitor; opposite charges of density

\sigma_p

appear on the slab faces. The dielectric constant K is given by: (A)

\dfrac{\sigma+\sigma_p}{\sigma}

(B)

\dfrac{\sigma}{\sigma-\sigma_p}

(C)

\dfrac{\sigma+\sigma_p}{\sigma_p}

(D)

\dfrac{\sigma}{\sigma_p}

(iii) An electric field E is established between the plates of an air-filled parallel plate capacitor with charges Q and

-Q

; V is the volume enclosed. The energy stored is: (A)

\varepsilon_0 E^2

(B)

\varepsilon_0 Q^2 E

(C)

\frac{1}{2}\varepsilon_0 E^2 V

(D)

\varepsilon_0 E Q V

. (iv)(a) Three capacitors A, B and M, each of capacitance C, are connected to a capacitor N of capacitance 2C and a battery as shown in the figure. If the charges on A and N are

Q_A

and

Q_N

respectively, then

Q_N/Q_A

is: (A)

\dfrac{1}{6}

(B)

\dfrac{1}{3}

(OR)

(iv)(b) A slab (area A and thickness d/2) of dielectric constant K is inserted in a parallel plate capacitor of plate area A and plate separation d. If C and

C_0

are the capacitances with and without the dielectric, then

C/C_0

is: (A)

\dfrac{K+1}{2K}

(B)

\dfrac{2K}{K+1}

(C)

\dfrac{K}{K-1}

(D)

\dfrac{K-1}{K}

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Part (a)Concept understanding — Capacitance of a Parallel Plate Capacitor

A parallel plate capacitor (two plates of area A, separation d, no dielectric) has capacitance $C_0=\dfrac{\epsilon_0A}{d}$ -- directly proportional to the plate area and inversely proportional to their separation. This is derived from the field inside the gap, $E=\dfrac{Q}{\epsilon_0A}$ (exactly twice a single charged sheet's field, since the two oppositely-charged plates' individual fields add inside the gap while cancelling to zero outside it), combined with $V=Ed$ . …

Part (b)Concept understanding — Dielectric Insertion Capacitance

Dielectric Insertion Capacitance – From Intuition to Precision

Imagine you have two metal plates facing each other, separated by air. You connect them to a battery. The plates get charged — one positive, one negative — and they store energy in the electric field between them. That's a capacitor.

Now, without disconnecting the battery, you slide a slab of some insulating material (like glass, plastic, or mica) between the plates. What happens? The battery pushes more charge onto the plates. The capacitor now stores more charge for the same voltage. Its capacitance has increased.

That increase — the extra capacitance contributed by the presence of the dielectric — is what we call dielectric insertion capacitance.

Note

The word "insertion" here simply means "the capacitance that appears because you inserted a dielectric." It's not a separate device; it's the change in capacitance due to the material.

Why does the capacitance increase?

The key is polarisation. Inside a dielectric, molecules are like tiny electric dipoles — they have a positive end and a negative end. In an electric field, these dipoles rotate to align with the field. The positive ends point toward the negative plate, and the negative ends point toward the positive plate.

This alignment creates a layer of bound charge on the surfaces of the dielectric, right next to the plates. This bound charge partially cancels the electric field inside the dielectric. But here's the crucial point: if the capacitor is connected to a battery (constant voltage), the battery responds by pushing more free charge onto the plates to restore the original field. More charge for the same voltage means higher capacitance.

If the capacitor is disconnected from the battery (constant charge), the dielectric reduces the voltage across the plates. Same charge, lower voltage — again, higher capacitance.

Watch out

A common mistake: thinking the dielectric "creates" extra charge out of nothing. It doesn't. The battery supplies the extra charge in the constant-voltage case. In the constant-charge case, the voltage drops — the capacitance formula $C = Q/V$ still gives a larger $C$ because $V$ is smaller.

The precise statement

Let’s define:

$C_0$ = capacitance of the capacitor with vacuum (or air) between the plates.
$\kappa$ (or $\varepsilon_r$ ) = dielectric constant (relative permittivity) of the material. For vacuum, $\kappa = 1$ . For most solids, $\kappa > 1$ (e.g., glass ~5–10, water ~80).

When you fill the entire space between the plates with a dielectric of constant $\kappa$ , the new capacitance is:

C = \kappa \, C_0

The dielectric insertion capacitance is the additional capacitance contributed by the dielectric:

C_{\text{insertion}} = C - C_0 = (\kappa - 1) C_0

C_{\text{insertion}} = (\kappa - 1) C_0

This is the extra capacitance you get purely because you inserted the dielectric. If $\kappa = 1$ (vacuum), $C_{\text{insertion}} = 0$ — no insertion effect.

What if the dielectric only partially fills the gap?

In real problems, the slab might not fill the entire space. Then the capacitor behaves like two capacitors in series (or parallel, depending on geometry). The insertion capacitance is no longer a simple multiple — you have to compute the effective capacitance using the appropriate combination rules.

But the core idea remains: the dielectric increases capacitance because its polarisation reduces the net field (or, equivalently, allows more charge at the same voltage).

Tip

For a parallel-plate capacitor with plate area $A$ , separation $d$ , and a dielectric of thickness $t$ inserted (leaving an air gap of $d-t$ ), the effective capacitance is:

$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{\kappa}}$ …

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(i) $C\to C/4$ ; (ii) $K=\frac{\sigma}{\sigma-\sigma_p}$ ; (iii) $U=\frac12\varepsilon_0E^2V$ ; (iv)(a) $Q_N/Q_A=6$ (bridge network -- N sits directly across the battery, P1/P2 are floating nodes); (iv)(b) $C/C_0=\frac{2K}{K+1}$ .

Part (a)

(i) Change of geometry

For a parallel-plate capacitor $C=\frac{\varepsilon_0A}{d}$ . Doubling the separation to $2d$ and halving the overlapping area to $A/2$ :

C'=\frac{\varepsilon_0(A/2)}{2d}=\frac14\cdot\frac{\varepsilon_0A}{d}=\frac{C}{4}\quad\text{(A)}.

(ii) Dielectric constant from charge densities

The free-charge field is $E_0=\frac{\sigma}{\varepsilon_0}$ . Induced polarisation charges $\sigma_p$ oppose it, so the net field inside is $E=\frac{\sigma-\sigma_p}{\varepsilon_0}$ . By definition

K=\frac{E_0}{E}=\frac{\sigma/\varepsilon_0}{(\sigma-\sigma_p)/\varepsilon_0}=\frac{\sigma}{\sigma-\sigma_p}\quad\text{(B)}.

(iii) Energy in terms of field and volume

U=\frac12CV^2=\frac12\frac{\varepsilon_0A}{d}(Ed)^2=\frac12\varepsilon_0E^2(Ad)=\frac12\varepsilon_0E^2V\quad\text{(C)},

so the energy density is $\frac12\varepsilon_0E^2$ .

(iv)(a) Capacitor network

The stored working here was based on the wrong topology -- once the actual network figure (previously missing) is drawn, it shows a bridge, not a simple series-parallel chain: capacitor A bridges the top rail between nodes P1 and P2; B connects P1 down to P3 (in series on the left); M connects P2 down to P4 (in series on the right); N bridges P3-P4, and the battery is connected directly across the SAME two nodes P3-P4 (in parallel with N).

Because N sits directly across the ideal battery, the potential difference across N is exactly the battery emf $V$ , regardless of A/B/M:

Q_N=(2C)\,V.

Nodes P1 and P2 are floating (each touches only two capacitors, no direct wire to the battery), so charge conservation applies at each: the charge flowing onto one capacitor's plate at that node must be supplied by the other capacitor at the same node. Taking $V_{P3}=V$ , $V_{P4}=0$ , and writing $V_1=V_{P1}$ , $V_2=V_{P2}$ :

\underbrace{C(V_1-V_2)}_{Q_A\text{ (P1 plate)}}+\underbrace{C(V_1-V)}_{Q_B\text{ (P1 plate)}}=0,\qquad -\underbrace{C(V_1-V_2)}_{Q_A\text{ (P2 plate)}}+\underbrace{C\,V_2}_{Q_M\text{ (P2 plate)}}=0.

Solving this pair: the second equation gives $V_1=2V_2$ ; substituting into the first gives $V_2=2V_1-V=4V_2-V$ , so $V_2=\dfrac{V}{3}$ and $V_1=\dfrac{2V}{3}$ .

The charge on A is then

Q_A=C(V_1-V_2)=C\left(\frac{2V}{3}-\frac{V}{3}\right)=\frac{CV}{3}.

$\frac{Q_N}{Q_A}=\frac{2CV}{CV/3}=6\quad\text{(D)}.$ …

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

Showing the 12 most recent of 26 on this concept.

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
$Q. The capacitance of a parallel plate capacitor ________ (increases/decreases) when a dielectric is inserted between the plates.$
›Reveal solutionSolution

Inserting a dielectric between the plates of a parallel plate capacitor increases its capacitance by a factor equal to the dielectric constant K.

For a parallel plate capacitor of plate area A and separation d, in vacuum $C_0 = \dfrac{\varepsilon_0 A}{d}$ . When a dielectric slab of dielectric constant K completely fills the gap, the dielectric gets polarised by the field between the plates, and the induced bound charge on its surfaces sets up a field that partly cancels the field of the free charges on the plates. For the same free charge Q, the field (and hence the potential difference V = Ed) between the plates is reduced. Since $C = Q/V$ , a sm …

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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 55/5/11 markMCQ
$C$
›Reveal solutionSolution
A conducting sheet has zero internal field, so it removes its own thickness $t$ from the effective gap, leaving $C' = \dfrac{\varepsilon_0 A}{d-t} > C$ . Option (B).

Let the plate area be $A$ and separation $d$ , so $C = \dfrac{\varepsilon_0 A}{d}$ . Insert a metal sheet of thickness $t\,(<d)$ that partly fills the gap.

Field inside a conductor is zero. The sheet develops induced charges on its two faces and carries no field within it, so it contributes nothing to the potential drop. Only the air gaps on either side of the sheet — of total thickness $d - t$ — sustain the field.

Effective separation shrinks. The capacitor behaves as if the plate gap were reduced from $d$ to $d - t$ :

$C' = \frac{\varepsilon_0 A}{d - t}.$

Compare with $C$ . Since $t > 0$ , we have $d - t < d$ , hence …
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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 2025Set ANNUAL1 markMCQ
$Q. If a dielectric is kept between two plates of capacitor, its capacitance :$
(a) increases
(b) decreases
(c) First increases then decreases
(d) none of these
›Reveal solutionSolution

Inserting a dielectric between the plates of a capacitor increases its capacitance by a factor equal to the dielectric constant $K$ (K > 1).

For a parallel plate capacitor, capacitance without dielectric is $C_0 = \varepsilon_0 A/d$ . When a dielectric slab of dielectric constant $K$ fills the gap, the dielectric partially cancels the field between the plates (due to induced polarization charges), so more charge can be stored for the …

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CBSE 2025Set ANNUAL1 markMCQ
$C$
(a) $C$
(b) $2C$
(c) $C/2$
(d) $C/4$
›Reveal solutionSolution

Inserting a dielectric of dielectric constant $k$ between the plates of a parallel-plate capacitor multiplies its capacitance by $k$ ; here $k=2$ gives $C'=2C$ .

Setup

For a parallel-plate capacitor with air (or vacuum) between the plates, plate area $A$ and separation $d$ :

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

…

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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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Part (a)

(i) Change of geometry

(ii) Dielectric constant from charge densities

(iii) Energy in terms of field and volume

(iv)(a) Capacitor network

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