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Physics · Electrostatic Potential and Capacitance

Effect of Dielectric on Capacitance

Filling the gap of a parallel plate capacitor with a dielectric of constant $K$ raises its capacitance to $C = K\varepsilon_0 A/d = KC_0$. NCERT Section 2.13 derives this single result, but the examinable subtlety lies in what is held fixed afterwards — charge or voltage. This note works through both cases, the partially-filled slab, and the conductor limit, all grounded in the NCERT relations and the way NEET keeps testing them.

Why capacitance rises

When a slab of dielectric is slid fully into the region between the plates, its molecules polarise in the applied field. As explained in dielectrics and polarisation, this polarisation is equivalent to two bound charged sheets — surface charge densities $+\sigma_P$ and $-\sigma_P$ — at the faces of the dielectric. These bound charges sit opposite the free plate charges and partly cancel them, so the net surface density driving the field falls from $\sigma$ to $(\sigma - \sigma_P)$.

A weaker net field means a smaller field $E$ inside the dielectric. Since capacitance is $C = Q/V$ and $V = Ed$, a reduction in the field for the same stored charge means a smaller voltage and therefore a larger capacitance. NCERT shows this works out cleanly to a single multiplicative factor: the capacitance becomes exactly $K$ times its vacuum value.

Figure 1 + + + + + + + + + − − − − − − − − − Dielectric K − − − − − − − − + + + + + + + + E = E₀/K d

The polarised slab develops bound charges that oppose the plate charges. The net field falls to $E_0/K$, raising the capacitance to $KC_0$.

NCERT derivation of C = KC₀

Start from vacuum. With free charge density $\sigma = Q/A$, the field is $E_0 = \sigma/\varepsilon_0$, the voltage is $V_0 = E_0 d$, and the capacitance is

$$C_0 = \frac{Q}{V_0} = \frac{\varepsilon_0 A}{d}$$

Insert the dielectric. The field becomes $E = (\sigma - \sigma_P)/\varepsilon_0$, so the voltage is $V = (\sigma - \sigma_P)\,d/\varepsilon_0$. For a linear dielectric the bound density is proportional to the free density, and NCERT writes $(\sigma - \sigma_P) = \sigma/K$ with $K > 1$. Hence

$$V = \frac{\sigma d}{K \varepsilon_0} = \frac{Q d}{K \varepsilon_0 A}, \qquad C = \frac{Q}{V} = \frac{K \varepsilon_0 A}{d}$$

Comparing with the vacuum value gives the master relation, NCERT Eq. (2.54):

$$\boxed{\,C = K C_0\,}$$

The dielectric constant $K$ is therefore defined as the factor by which the capacitance increases when the dielectric is inserted fully. The product $\varepsilon = \varepsilon_0 K$ is the permittivity of the medium — exactly what NEET 2020 asked you to compute from $C_m = K C_0$.

The two cases: Q-const vs V-const

The capacitance rises to $KC_0$ no matter what. Everything else — voltage, charge, field, stored energy — depends entirely on a single circuit fact: was the battery still connected when the slab went in? This is the most-tested distinction in the entire topic.

Case A — battery disconnected (Q held constant)

The capacitor is charged, then unplugged, then the slab is inserted. Isolated plates cannot lose or gain charge, so $Q$ is frozen. With $C$ rising to $KC_0$:

$$V = \frac{Q}{C} = \frac{Q}{KC_0} = \frac{V_0}{K}, \quad E = \frac{E_0}{K}, \quad U = \frac{Q^2}{2C} = \frac{U_0}{K}$$

Voltage, field and energy all drop by a factor $K$. The lost energy is the work the dielectric does as it is pulled inward — the slab is sucked into the gap, lowering the system energy.

Case B — battery connected (V held constant)

The battery stays attached, pinning the voltage at its EMF. With $C$ rising to $KC_0$ and $V$ fixed:

$$Q = CV = KC_0 V_0 = K Q_0, \quad E = \frac{V_0}{d} = E_0, \quad U = \tfrac{1}{2}CV^2 = K U_0$$

Charge and energy rise by a factor $K$; the field stays the same because both $V$ and $d$ are unchanged. The battery pushes extra charge $K Q_0$ onto the plates and supplies the extra stored energy.

Build the base first

All of this rests on the vacuum result $C_0=\varepsilon_0 A/d$. Revisit the parallel plate capacitor derivation if the geometry feels shaky.

NEET Trap

Q-constant and V-constant give OPPOSITE energy changes

The single most common error: assuming the energy always rises when a dielectric is inserted. It does not. With the battery disconnected (Q fixed) the energy falls to $U_0/K$; with the battery connected (V fixed) it rises to $K U_0$. Read the question for the word "disconnected" or "still connected" before touching the energy formula.

Q fixed → use $U = Q^2/2C$ → energy ↓.   V fixed → use $U = \tfrac12 CV^2$ → energy ↑.

NEET Trap

Capacitance always rises — that part never flips

Whatever is held constant, $C$ becomes $KC_0$ in both cases because $C = K\varepsilon_0 A/d$ depends only on geometry and the medium, not on the circuit. Only $V$, $Q$, $E$ and $U$ depend on the connection. Do not let a "battery disconnected" stem trick you into thinking the capacitance is unchanged.

$C \to KC_0$ in BOTH cases. The connection only decides which of $V$ and $Q$ is frozen.

Master comparison table

The table below is the spine of every dielectric problem. Memorise the two columns — which quantity is constant, and which way each of the others moves.

Quantity Battery disconnected (Q constant) Battery connected (V constant)
Capacitance C KC₀  (rises ×K) KC₀  (rises ×K)
Charge Q Q₀  (fixed) KQ₀  (rises ×K)
Voltage V V₀/K  (falls) V₀  (fixed)
Field E E₀/K  (falls) E₀  (fixed)
Energy U U₀/K  (falls) KU₀  (rises ×K)
Energy source slab pulled in; system loses U battery supplies extra Q and U

Partially-filled capacitor

If the dielectric does not fill the whole gap — a slab of constant $K$ and thickness $t$ (with $t < d$) sits parallel to the plates — the gap behaves as an air layer of length $(d - t)$ in series with a dielectric layer of length $t$. Following NCERT Example 2.8, the dielectric layer acts like an air gap of reduced length $t/K$, so the effective separation is $(d - t + t/K)$ and

$$C = \frac{\varepsilon_0 A}{d - t + \dfrac{t}{K}}$$

This one formula contains both limits. Setting $t = d$ recovers $C = K\varepsilon_0 A/d = KC_0$ (fully filled); setting $t = 0$ recovers $C = \varepsilon_0 A/d = C_0$ (empty). NEET 2025 extended this to two stacked slabs using the same series logic.

Figure 2 air gap (d − t) dielectric K, t t d

A slab of thickness $t$ leaves an air layer $(d-t)$. The dielectric counts as only $t/K$ of effective gap, so $C = \varepsilon_0 A/(d - t + t/K)$.

Worked Example

A slab of dielectric constant $K$ has thickness $\tfrac{3}{4}d$ and the same area as the plates. By what factor does the capacitance change? (NCERT Example 2.8)

Here $t = \tfrac34 d$, so the effective gap is $d - \tfrac34 d + \tfrac{3d}{4K} = \tfrac{d}{4} + \tfrac{3d}{4K} = \dfrac{d(K + 3)}{4K}$.

Therefore $C = \dfrac{\varepsilon_0 A}{d(K+3)/4K} = \dfrac{4K}{K+3}\,C_0$. The voltage falls by the factor $(K+3)/4K$ at fixed charge, and the capacitance rises by $4K/(K+3)$.

Conducting slab (K → ∞)

A conductor has no internal field, so it behaves like a dielectric with $K \to \infty$. In the partial-fill formula the term $t/K \to 0$, leaving

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

A conducting slab of thickness $t$ simply removes a length $t$ of the gap: the effective separation shrinks from $d$ to $(d - t)$, raising the capacitance. Note that this result is independent of where the slab sits within the gap, since a conductor is an equipotential region.

NEET Trap

Partial fill ≠ simply multiply by K

A slab thinner than the gap does not give $C = KC_0$. Students who skip the $(d - t + t/K)$ denominator and write $KC_0$ overshoot the answer badly. Use $KC_0$ only when the dielectric fills the gap completely, i.e. $t = d$.

Full fill: $C = KC_0$.   Partial fill: $C = \varepsilon_0 A/(d - t + t/K)$.

Quick Recap

Effect of dielectric in one screen

  • Full dielectric fill raises capacitance to $C = K\varepsilon_0 A/d = KC_0$ — always, in every case ($K>1$).
  • Battery disconnected (Q fixed): $V \to V_0/K$, $E \to E_0/K$, $U \to U_0/K$.
  • Battery connected (V fixed): $Q \to KQ_0$, $E$ unchanged, $U \to KU_0$.
  • Energy moves in opposite directions in the two cases — read whether the battery is removed.
  • Partial slab of thickness $t$: $C = \varepsilon_0 A/(d - t + t/K)$.
  • Conducting slab ($K\to\infty$): $C = \varepsilon_0 A/(d - t)$ — gap effectively shortened.

NEET PYQ Snapshot — Effect of Dielectric on Capacitance

Dielectric effects are tested almost every year — directly through $C = KC_0$ and through partial-fill geometry.

NEET 2020

The capacitance of a parallel plate capacitor with air as medium is 6 µF. With a dielectric medium introduced, the capacitance becomes 30 µF. The permittivity of the medium is ($\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$):

  1. $1.77 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$
  2. $0.44 \times 10^{-10}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$
  3. $5.00\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$
  4. $0.44 \times 10^{-13}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$
Answer: (2)

From $C_m = K C_0$, $K = 30/6 = 5$. Then $\varepsilon = \varepsilon_0 K = 8.85\times10^{-12}\times5 = 0.44\times10^{-10}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$.

NEET 2024

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then: A. the charge stored increases; B. the energy stored decreases; C. its capacitance increases; D. the ratio of charge to potential remains the same; E. the product of charge and voltage increases.

  1. A, B and E only
  2. A, C and E only
  3. B, D and E only
  4. A, B and C only
Answer: (2)

Battery connected → $V$ fixed. Reducing $d$ raises $C = \varepsilon_0 A/d$ (C true); $Q = CV$ rises (A true); $U = \tfrac12 CV^2$ rises and $QV = CV^2$ rises (E true, B false). The ratio $Q/V = C$ is not constant since $C$ changes (D false). The same V-constant reasoning governs a dielectric insertion at fixed voltage.

NEET 2025

The plates of a parallel plate capacitor are separated by $d$. Two slabs of dielectric constants $K_1$ and $K_2$ with thicknesses $\tfrac38 d$ and $\tfrac{d}{2}$ are inserted, making the capacitance twice the empty value. If $K_1 = 1.25\,K_2$, the value of $K_1$ is:

  1. 1.33
  2. 2.66
  3. 2.33
  4. 1.60
Answer: (2)

Series of layers: $C = \varepsilon_0 A/(t_1/K_1 + t_2/K_2 + t_3/K_3)$ with $t_1=\tfrac38 d$, $t_2=\tfrac{d}{2}$, remaining air gap $t_3=\tfrac{d}{8}$, $K_3=1$. With $K_2 = K_1/1.25$ and $C = 2C_0$, solving gives $K_1 = 8/3 = 2.66$ — a direct extension of the partial-fill formula.

FAQs — Effect of Dielectric on Capacitance

The questions NEET aspirants ask most about dielectrics and the two charging cases.

What happens to capacitance when a dielectric is inserted?
Capacitance always rises by the factor K. A fully inserted dielectric of constant K changes C₀ = ε₀A/d to C = Kε₀A/d = KC₀. Because K > 1 for every real dielectric, the capacitance increases in every case, whether the battery is connected or disconnected.
Why does the energy stored fall in one case but rise in the other?
It depends on what is held fixed. With the battery disconnected, charge Q is fixed and U = Q²/2C, so when C rises K times the energy drops to U₀/K. With the battery connected, voltage V is fixed and U = ½CV², so when C rises K times the energy rises to KU₀ — the extra energy is supplied by the battery.
What is the capacitance of a partially-filled capacitor?
If a dielectric slab of constant K and thickness t (t < d) is inserted parallel to the plates, C = ε₀A / (d − t + t/K). When t = d this reduces to Kε₀A/d, and when t = 0 it returns to ε₀A/d. The slab behaves like an air gap of reduced effective length t/K.
How does the electric field change when a dielectric is inserted at constant charge?
With charge held fixed (battery disconnected) the field inside the dielectric falls to E = E₀/K. The polarised dielectric sets up bound surface charges that partly cancel the plate charge, reducing the net surface density and hence the field. The potential difference V = Ed also drops to V₀/K.
What happens if a conducting slab is inserted instead of a dielectric?
A conductor behaves like a dielectric with K → ∞, so the t/K term vanishes. The partial-fill formula gives C = ε₀A/(d − t), meaning a conducting slab of thickness t simply reduces the effective plate separation from d to (d − t), raising the capacitance.
Does inserting a dielectric change the charge on a battery-connected capacitor?
Yes. When the battery stays connected, V is held constant. Since Q = CV and C rises to KC₀, the charge rises to Q = KQ₀. The extra charge flows from the battery onto the plates while the voltage is pinned at the battery EMF.
Related Topics
Electrostatic Potential and Capacitance Back to the full chapter hub. Dielectrics and Polarisation Where the bound charges and factor K come from. Parallel Plate Capacitor The vacuum result C₀ = ε₀A/d this builds on. Capacitors and Capacitance Definition C = Q/V and its meaning. Energy Stored in a Capacitor Why Q-const and V-const energies differ. Combination of Capacitors Series logic behind the partial-fill formula.