(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separatio

Question

A

(b) Two air-filled capacitors of capacitances

C_1

and

C_2

are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant

K

is inserted between the plates of each capacitor. How will the

(i) charge on each capacitor and

(ii) energy stored in the capacitor be affected after the slab is introduced?

(OR)

(a) An electric field

E

is established across the ends of a cylindrical conductor of length

L

and area of cross-section

A

. Discuss how electrons attain an average velocity, independent of time. Hence obtain a relation between the current in the conductor and this average velocity of electrons.

(b)

(i) This average velocity is found to be a few mm/s for currents in the range of a few amperes. How then is current established almost the instant a circuit is closed?

(ii) Two copper wires having their radii in the ratio

3:2

are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

CBSECBSE Class XII Board 2026Subjective· 5mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)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:

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

Part (b)Concept understanding — Drift Velocity

Drift Velocity: The Slow March of Electrons

Electrons in a metal are always moving — but randomly. At room temperature they zip around at roughly $10^6$ m/s, colliding with the lattice ions every few trillionths of a second. Without an electric field this motion cancels out: for every electron heading left another heads right, so the net velocity is zero.

Apply a battery and the field gives every electron a tiny, steady push in one direction. Between collisions the electron accelerates only briefly before smashing into an ion and losing its directed motion. What survives is a very small average velocity along the field — the drift velocity.

Note

The random thermal speed is about $10^5$ m/s, but the drift velocity is only about $10^{-4}$ m/s — about a billion times slower. An electron drifts slower than a snail, yet a lamp lights instantly, because the electric field (not the electrons) propagates at nearly the speed of light and starts every electron drifting almost at once.

The precise definition

Drift velocity ( $v_d$ ) is the average velocity acquired by the charge carriers in a conductor under an applied electric field:

v_d = \frac{eE\tau}{m}

where:

$e$ = electron charge ( $1.6 \times 10^{-19}$ C)
$E$ = electric field inside the conductor (V/m)
$\tau$ = average relaxation time — the mean time between collisions (s)
$m$ = electron mass ( $9.1 \times 10^{-31}$ kg)

v_d = \frac{eE\tau}{m}

Linking to current

Drift velocity connects the microscopic motion of electrons to the current an ammeter reads:

I = n e A v_d

where $n$ is the free-electron number density and $A$ the cross-sectional area. A larger $v_d$ means more current, but $v_d$ stays tiny because $\tau$ is tiny (about $10^{-14}$ s in copper).

Tip

For a copper wire carrying $1$ A with area $1$ mm² and $n \approx 8.5 \times 10^{28}$ m $^{-3}$ :

$v_d = \frac{I}{neA} \approx \frac{1}{(8.5\times10^{28})(1.6\times10^{-19})(10^{-6})} \approx 7 \times 10^{-5}\ \text{m/s}$ …

Why this formula?

Drift Velocity: Why the Formula Holds

Let's build this from first principles — understanding why electrons drift the way they do, not just memorizing the formula.

1. The Core Idea: What is Drift Velocity?

In a conductor, free electrons are constantly moving randomly (thermal motion, speeds ~ $10^5$ m/s). Without an electric field, their net displacement is zero — they're like a swarm of bees buzzing in all directions.

When we apply an electric field $\vec{E}$ , it gently nudges each electron in the opposite direction (since electrons are negatively charged). This small, steady net velocity superimposed on the random motion is drift velocity ( $v_d$ ).

Key insight: Drift velocity is not the speed of individual electrons — it's the average velocity of the entire electron cloud.

2. The Derivation: Step by Step

Step 1: Force on a single electron

An electron of charge $-e$ in an electric field $\vec{E}$ experiences:

\vec{F} = -e \vec{E}

The magnitude of acceleration (opposite to $\vec{E}$ ) is:

a = \frac{F}{m} = \frac{eE}{m}

where $m$ is the electron's mass.

Step 2: What happens between collisions?

Electrons don't accelerate forever — they keep colliding with atoms/ions in the metal lattice. Let the average time between collisions be $\tau$ (relaxation time). Just after a collision an electron's velocity is essentially random (zero average in the field direction); it then accelerates for time $\tau$ before the next collision.

Step 3: Drift velocity

Averaging the field-driven velocity over the relaxation time $\tau$ gives the net drift:

\boxed{v_d = \frac{eE\tau}{m}}

Here $\tau$ is the average time since the last collision, so this expression already averages over electrons at every stage between collisions — it is the standard result used in the NCERT treatment.

3. Connecting to Current: The Big Picture

Drift velocity directly gives us current density $J$ :

J = n e v_d

where $n$ = number of free electrons per unit volume. …

Start your 14-day free trial to unlock the full solution →

$C=\dfrac{\varepsilon_0 A}{d}$ ; inserting a dielectric with the battery connected multiplies both charge and energy of each capacitor by $K$ . OR: $I=neAv_d$ ; current appears instantly because the field travels near light speed; series drift-velocity ratio $=4:9$ .

Part (a) — parallel-plate capacitor and dielectric insertion

Derivation of $C$ . For plates of area $A$ separated by $d$ with charge $\pm Q$ , the field between them (air) is

E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0 A},

so the potential difference is $V=Ed=\dfrac{Qd}{\varepsilon_0 A}$ and

C=\frac{Q}{V}=\frac{\varepsilon_0 A}{d}.

Dielectric slab inserted with the battery still connected (parallel $C_1,C_2$ ). In parallel, each capacitor stays at the battery voltage $V$ ; inserting the slab raises each capacitance to $KC$ .

(i) Charge: $Q'=KC\,V=KQ$ — the charge on each capacitor increases by the factor $K$ (the battery supplies the extra charge).
(ii) Energy: $U'=\tfrac12(KC)V^2=K\big(\tfrac12 CV^2\big)=KU$ — the stored energy of each capacitor also increases by $K$ . …

Start your 14-day free trial to unlock the full solution →

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 38 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Drift velocity Vd varies with the intensity of electric field E as per the relation$
(a) Vd is proportional to E^2
(b) Vd is proportional to 1/E
(c) Vd is proportional to sqrt(E)
(d) Vd is proportional to E
›Reveal solutionSolution

Drift velocity is the (small) average velocity electrons gain between collisions due to the electric field, and it comes out directly proportional to E.

When an electric field E is applied to a conductor, each free electron experiences a force F = eE, giving it an acceleration a = eE/m between collisions with the lattice ions. If tau is the average time between collisions (relaxation time), the average extra velocity gained (the drift velocity) is

vd = a * tau = (eE/m) * tau = (e*tau/m) * E

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. If the current flowing in a copper wire be allowed to flow in another copper wire of same length but of doubled the radius then what will be the effect on the drift velocity of the electron?$
›Reveal solutionSolution

For the same current, $v_d \propto 1/A$ , and doubling the radius quadruples the cross-sectional area.

Current is related to drift velocity by $I = nAev_d$ , so for the same current $I$ (and the same material, hence the same $n$ ), $v_d = \dfrac{I}{nAe} \propto \dfrac{1}{A}$ . If the radius is doubled, the cross-sectional area $A=\pi r^2$ becomes 4 times larger. So the drift velocity becomes

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. State Ohm's law in terms of current density, specific conductance and electric field intensity.$
›Reveal solutionSolution

Microscopic Ohm's law: current density J = σE (σ = conductivity, E = field).

The usual Ohm's law is V = IR. In microscopic (vector) form, it relates the current density J (current per unit cross-sectional area) to the electric field E inside the conductor through the material's specific conductance (conductivity) σ:

J = σ E.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set SEM31 markMCQ
$Q. Which of the following statement(s) is/are true ? A potential difference of V is applied at the two ends of a conductor of length l and area of cross-section A. Statement I : When potential difference is doubled, current density also gets doubled. Statement II : When potential difference is doubled, drift velocity gets halved. Statement III : When area of cross-section is doubled, current density decreases.$
(a) I and II are true
(b) Only I is true
(c) Only III is true
(d) II and III are true
›Reveal solutionSolution

Doubling V doubles E, so J = σE and v_d = μE both double — Statement I true, Statement II (drift velocity halved) false. J = V/(ρl) is independent of area, so Statement III (J decreases when A doubles) is also false. Only I is true → option (b).

Statement I: J = σE and E = V/l, so doubling V doubles E and hence doubles the current density J. TRUE.

Statement II: drift velocity v_d = (eE/m)τ ∝ E ∝ V. Doubling V doubles v_d, it does not halve it. FALSE.

…

Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. The relation between drift velocity v of free electrons in conductor in electric conduction and potential difference V between ends of conductor is (A) proportional to V (B) inversely proportional to V (C) proportional to V^2 (D) inversely proportional to V^2$
›Reveal solutionSolution

Drift velocity is directly proportional to the potential difference V.

In a conductor of length L across which a potential difference V is applied, the electric field is E = V/L. Free electrons acquire a drift velocity

$v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$ …

Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. If the length of a conductor is doubled while keeping the potential difference across it constant, then the drift velocity of electron will (A) remain the same (B) be double (C) be halved (D) increase fourfold$
›Reveal solutionSolution

Doubling the length at constant V halves the drift velocity.

The drift velocity is

$v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$ …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 $d$ :

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

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. A thick wire is stretched so that its length becomes two times. What is the ratio of change in resistance of the wire to the initial resistance of the wire?$
(i) 2 : 1
(ii) 4 : 1
(iii) 3 : 1
(iv) 1 : 4
›Reveal solutionSolution

New resistance is 4 times the old, so the change is 3 times the original: ratio 3 : 1.

Resistance $R = \rho L/A$ . Stretching keeps the volume $AL$ constant, so if length doubles ( $L \to 2L$ ) the area halves ( $A \to A/2$ ). Then $R' = \rho (2L)/(A/2) = 4\rho L/A = 4R$ . The change in …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Calculate the amount of charge flowing in 2 minutes in a wire of resistance 10 ohm when a potential difference of 20 volts is applied between its ends.$
(i) 120 C
(ii) 240 C
(iii) 20 C
(iv) 4 C
›Reveal solutionSolution

Q = It = (V/R) x t = 2 A x 120 s = 240 C.

…

Start your 14-day free trial to see the full solution →

Q.(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area $A$ and plate separation $d$ with air present between the plates.

Part (a)Concept understanding — Dielectric Insertion Capacitance

Dielectric Insertion Capacitance – From Intuition to Precision

Why does the capacitance increase?

The precise statement

What if the dielectric only partially fills the gap?

Part (b)Concept understanding — Drift Velocity

Drift Velocity: The Slow March of Electrons

The precise definition

Linking to current

Why this formula?

Drift Velocity: Why the Formula Holds

1. The Core Idea: What is Drift Velocity?

2. The Derivation: Step by Step

Step 1: Force on a single electron

Step 2: What happens between collisions?

Step 3: Drift velocity

3. Connecting to Current: The Big Picture

Parallel-plate capacitor / drift velocity

Part (a) — capacitor + dielectric in parallel

Part (a) — parallel-plate capacitor and dielectric insertion

Q.(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area AAA and plate separation ddd with air present between the plates.

Part (a) — parallel-plate capacitor and dielectric insertion

🎓Unlock everything free for 14 days

Q.(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area $A$ and plate separation $d$ with air present between the plates.

Unlock everything free for 14 days