A parallel plate capacitor consists of two conducting plates kept parallel to each other at a distance. When the capacitor is charged, the charge resides on the inner surfaces of the plates and an electric field is set up between them, storing electrostatic energy. Three identical conducting plates $P_1$, $P_2$ and $P_3$, each of area $L^2$, are held parallel and equidistant (separation $d$) from each other. The space between $P_1$ and $P_2$ and between $P_2$ and $P_3$ is completely filled with mica sheets of dielectric constant $K$. Plate $P_2$ is connected to point A and the other plates $P_1$ and $P_3$ are connected to point B. Point A is at a positive potential with respect to B, and the potential difference between A and B is $V$. (i) The capacitance of the system between A and B will be: (ii) The charge on plate $P_1$ is: (iii) The electric field in the region between $P_1$ and $P_2$ is: (iv)(a) The separation between the plates of the same area ($L^2$) of a parallel plate air capacitor having capacitance equal to that of this system will be: OR (iv)(b) If the source of potential difference applied between A and B is removed, and then A and B are connected by a conducting wire, the net charge on the system will be: ![Figure — 55/4/1 Q29](/figures/pyq/2025-55-4-1/fig-q29.svg)

A parallel plate capacitor consists of two conducting plates kept parallel to each other at a distance. When t

Question

Figure — Figure — 55/4/1 Q29 — FigureFigure — 55/4/1 Q29

P_1

(i) The capacitance of the system between A and B will be:

(ii) The charge on plate

P_1

is:

(iii) The electric field in the region between

P_1

and

P_2

is: (iv)(a) The separation between the plates of the same area (

L^2

) of a parallel plate air capacitor having capacitance equal to that of this system will be:

(OR)

(iv)(b) If the source of potential difference applied between A and B is removed, and then A and B are connected by a conducting wire, the net charge on the system will be:

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Part (a)Concept understanding — Capacitors in Series and Parallel

In a SERIES combination, every capacitor carries the SAME charge Q, and the equivalent capacitance obeys $\dfrac{1}{C_S}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots$ -- always SMALLER than the smallest individual capacitor, and used to divide a high voltage safely across several capacitors (the smallest-capacitance capacitor in the chain always bears the largest share of the voltage). …

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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The three plates form two identical dielectric capacitors in parallel: (i) $C=\frac{2\varepsilon_0KL^2}{d}$ ; (ii) charge on $P_1$ is $\frac{\varepsilon_0KL^2V}{d}$ ; (iii) field $=V/d$ ; (iv)(a) equivalent air-gap $=d/2K$ ; (iv)(b) net charge after shorting A–B $=0$ .

Part (a)

Setting up the system

Plate $P_2$ is connected to A (potential $+V$ ); plates $P_1$ and $P_3$ are connected to B (potential 0). Thus $P_2$ is the common positive plate of two capacitors, each with area $L^2$ , separation $d$ and mica of dielectric constant $K$ . Because $P_1$ and $P_3$ share node B, the two capacitors are in parallel.

Each individual capacitance:

C_0=\frac{\varepsilon_0KL^2}{d}.

(i) Capacitance between A and B

C=C_0+C_0=\frac{2\varepsilon_0KL^2}{d}.

(ii) Charge on plate $P_1$

$P_1$ is one plate of the $P_1$ – $P_2$ capacitor, across which the p.d. is $V$ :

Q_1=C_0V=\frac{\varepsilon_0KL^2V}{d}.

(The magnitude of the charge on $P_1$ ; its sign is negative. $P_2$ carries twice this in total.)

(iii) Electric field between $P_1$ and $P_2$

The gap $d$ has the full applied p.d. $V$ across it, so the uniform field is

E=\frac{V}{d},

directed from $P_2$ (high) to $P_1$ (low).

(iv)(a) Equivalent air capacitor

An air-filled ( $K=1$ ) capacitor of area $L^2$ and separation $d'$ equal in capacitance requires …

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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 21 on this concept.

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 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
$Q. In parallel combination of condensers which quantity remains same for each condenser? (A) Charge (B) Energy (C) Potential difference (D) Capacity$
›Reveal solutionSolution

In a parallel combination, every capacitor shares the same two terminals, so the potential difference across each is identical.

When capacitors are connected in parallel, their plates are joined to the same pair of nodes. The voltage across each capacitor therefore equals the common terminal voltage V.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Determine the equivalent capacitance between the points A and B in the adjoining figure.$
›Reveal solutionSolution

First combine the two parallel capacitors, then combine that result in series with the 6 pF capacitor.

Step 1 — parallel combination: the 1 pF and 2 pF capacitors are connected in parallel between the two junction nodes:

$C_{parallel} = 1 + 2 = 3\ \text{pF}$

Step 2 — series combination: this 3 pF combination is in series with the 6 pF capacitor between A and B:

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Three capacitors each having capacitance C are connected in series. Their equivalent capacitance is$
(a) 3C
(b) 3/C
(c) C/3
(d) 1/3C
›Reveal solutionSolution

Capacitors in series combine like resistors in parallel: 1/C_eq = 1/C1 + 1/C2 + 1/C3, so three equal capacitors C give C_eq = C/3.

For capacitors in series, the same charge Q flows onto each, and the total voltage is the sum of individual voltages, which leads to:

1/C_eq = 1/C + 1/C + 1/C = 3/C

C_eq = C/3

…

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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 mark
$Q. In which order capacitors are connected to increase resultant capacitance?$
›Reveal solutionSolution
Parallel combination increases equivalent capacitance; series combination decreases it.

For capacitors $C_1, C_2, C_3, \ldots$ connected:

In series: $\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + \ldots$ — the equivalent capacitance is always smaller than the smallest individual capacitance (because effectively the plate separation increases / the charge stored is the same on each but the total voltage adds up). …
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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 2025Set ANNUAL1 mark
$Q. What is the equivalent capacitance between A and B ?$
›Reveal solutionSolution
This 5-capacitor network is a balanced capacitor bridge (all five are equal, 4 μF each), so the diagonal 'bridge' capacitor carries no charge and can be removed; what's left is two 2 μF series paths in parallel, giving 4 μF.

Label the apex of the triangle P and the mid-point node of the two base capacitors Q. The network is:

$A\!-\!P$ : 4 μF (left slant)

$P\!-\!B$ : 4 μF (right slant)

$A\!-\!Q$ : 4 μF (left half of base)

$Q\!-\!B$ : 4 μF (right half of base)

$P\!-\!Q$ : 4 μF (the bridging capacitor)

This is exactly a Wheatstone-bridge-like arrangement. A capacitor bridge is balanced (no charge on the bridge arm, hence no potential difference across it) when

$\dfrac{C_{AP}}{C_{AQ}} = \dfrac{C_{PB}}{C_{QB}}$

…
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The ratio between resultant capacitances Cs and Cp of two equal capacitors in series combination and parallel combination respectively is$
(a) 1
(b) 1/2
(c) 1/4
(d) 1/8
›Reveal solutionSolution

Two equal capacitors C in series give C_s = C/2; in parallel they give C_p = 2C; the ratio C_s/C_p = 1/4.

Let each capacitor have capacitance C.

Series combination: for two capacitors in series, 1/C_s = 1/C + 1/C = 2/C, so C_s = C/2.

Parallel combination: for two capacitors in parallel, capacitances simply add: C_p = C + C = 2C.

Now form the ratio:

C_s/C_p = (C/2)/(2C) = 1/4

…

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CBSE 2024Set ANNUAL1 mark
$Q. Two capacitors of capacities 5μF and 10μF respectively are connected in series. Calculate the resultant capacity of the combination.$
›Reveal solutionSolution

Reciprocal-sum rule for capacitors in series.

$\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2} = \frac{1}{5}+\frac{1}{10} = \frac{2}{10}+\frac{1}{10} = \frac{3}{10}$

$C_{eq} = \frac{10}{3} \approx 3.33\ \mu\text{F}$ …

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CBSE 2023Set F1 markMCQ
$Q. Three capacitors of capacitance 6 μF are available. The minimum and maximum capacitances obtained are (A) 3 μF, 12 μF (B) 2 μF, 12 μF (C) 2 μF, 18μF (D) 6 μF, 18 μF$
›Reveal solutionSolution

All in series ⇒ 2 μF (minimum); all in parallel ⇒ 18 μF (maximum).

For three equal capacitors of $C = 6\,\mu\text{F}$ :

…

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Part (a)

Setting up the system

(i) Capacitance between A and B

(ii) Charge on plate P1P_1P1​

(iii) Electric field between P1P_1P1​ and P2P_2P2​

(iv)(a) Equivalent air capacitor

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(ii) Charge on plate $P_1$

(iii) Electric field between $P_1$ and $P_2$

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