A 200 $\mu$F parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the (i) capacitance, (ii) electric field between the plates, (iii) energy density of the capacitor will change ?

A 200 μF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains

Question

\mu

(i) capacitance,

(ii) electric field between the plates,

(iii) energy density of the capacitor will change ?

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★

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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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Key idea: With the source still connected, the voltage $V=100$ V stays fixed across the WHOLE gap (air + slab in series). Any change in geometry or dielectric alters the capacitance $C$ ; the field then splits between the air gap and the slab according to continuity of $D=\varepsilon_0 E_{\text{air}}=\varepsilon_0 K E_{\text{diel}}$ , not by applying the full $V$ to either segment alone.

Step 1 – Initial state

$C_0 = 200\ \mu\text{F}$ , $d_0 = 5\ \text{mm}$ , $V = 100\ \text{V}$ .

Initial (uniform) field: $E_0 = V/d_0 = 100/(5\times10^{-3}) = 2\times10^4\ \text{V/m}$ .

Step 2 – After doubling the gap and inserting the slab

New total gap $d' = 10\ \text{mm}$ ; a slab of thickness $t=5\ \text{mm}$ , $K=10$ fills half of it, leaving a $5\ \text{mm}$ air gap. Effective gap for capacitance:

d'-t+\frac{t}{K}=10-5+\frac{5}{10}=5.5\ \text{mm}.

Since $\varepsilon_0 A = C_0 d_0 = (200\ \mu\text{F})(5\ \text{mm})$ :

C' = \frac{(200)(5)}{5.5}\ \mu\text{F} = \frac{2000}{11} \approx 181.8\ \mu\text{F}\quad(\text{decreases from }200\ \mu\text{F}).

Step 3 – Field in each region

$V=100$ V is shared between the air gap and the slab (both $5$ mm thick). With $D$ continuous across the boundary, $E_{\text{air}}=K\,E_{\text{slab}}$ , and $E_{\text{air}}t+E_{\text{slab}}t=V$ :

E_{\text{slab}}(K+1)t=V\ \Rightarrow\ E_{\text{slab}}=\frac{100}{11\times5\times10^{-3}}\approx1.82\times10^{3}\ \text{V/m},\qquad E_{\text{air}}=10\,E_{\text{slab}}\approx1.82\times10^{4}\ \text{V/m}.

Both are less than the original uniform $E_0=2\times10^4$ V/m.

Step 4 – Energy density ( $u=\tfrac12\varepsilon E^2$ )

Original: $u_0=\tfrac12\varepsilon_0(2\times10^4)^2\approx1.77\times10^{-3}\ \text{J/m}^3$ . …

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The capacitor stays connected, so $V = 100\ \text{V}$ is fixed. Doubling the gap to $10\ \text{mm}$ with a $5\ \text{mm}$ , $K=10$ slab gives an effective gap of $5.5\ \text{mm}$ : all three decrease - capacitance $200\to181.8\ \mu\text{F}$ , the field in the air gap $2\times10^{4}\to1.82\times10^{4}\ \text{V m}^{-1}$ (and $10\times$ smaller inside the slab), and the energy density falls correspondingly.

Setup

Initially $C_0 = \dfrac{\varepsilon_0 A}{d} = 200\ \mu\text{F}$ with $d = 5\ \text{mm}$ , charged to $V = 100\ \text{V}$ and left connected, so $V$ stays $100\ \text{V}$ . The gap is doubled to $d' = 10\ \text{mm}$ ; a slab of thickness $t = 5\ \text{mm}$ , $K = 10$ fills half of it, leaving a $5\ \text{mm}$ air gap.

(i) Capacitance - decreases

For a slab of thickness $t$ in a gap $d'$ : $\;C' = \dfrac{\varepsilon_0 A}{\,d' - t + t/K\,}$ .

d'-t+\frac{t}{K} = 10 - 5 + \frac{5}{10} = 5.5\ \text{mm}.

Since $\varepsilon_0 A = C_0 d = (200\ \mu\text{F})(5\ \text{mm})$ ,

C' = \frac{(200)(5)}{5.5}\ \mu\text{F} = \frac{2000}{11} \approx 181.8\ \mu\text{F}\ \ (\text{decreases from }200\ \mu\text{F}).

(ii) Electric field - decreases (and becomes non-uniform)

$V = 100\ \text{V}$ is shared between the air gap and slab. With $\vec D$ continuous, $E_{\text{air}} = K\,E_{\text{slab}}$ , and $E_{\text{air}}t + E_{\text{slab}}t = V$ :

E_{\text{slab}}(K+1)t = V \Rightarrow E_{\text{slab}} = \frac{100}{11\times5\times10^{-3}} \approx 1.82\times10^{3}\ \text{V m}^{-1},\quad E_{\text{air}} = 10\,E_{\text{slab}} \approx 1.82\times10^{4}\ \text{V m}^{-1}.

Compared with the original uniform $E_0 = V/d = 2\times10^{4}\ \text{V m}^{-1}$ , the air-gap field decreases slightly and the field inside the dielectric is $K = 10$ times smaller still.

(iii) Energy density - decreases

$u = \tfrac12 \varepsilon E^2$ . Originally $u_0 = \tfrac12\varepsilon_0 E_0^2 = \tfrac12(8.85\times10^{-12})(2\times10^{4})^2 \approx 1.77\times10^{-3}\ \text{J m}^{-3}$ . Now: …

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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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Setup

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

…

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Formula

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…

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…

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Setup

(i) Capacitance - decreases

(ii) Electric field - decreases (and becomes non-uniform)

(iii) Energy density - decreases

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