Capacitor with dielectric inserted — how does capacitance change

Question

A parallel plate capacitor is connected to a battery and has a capacitance $C_0$ . A dielectric slab of dielectric constant $K = 3$ is fully inserted between the plates while the battery remains connected. How does the capacitance, charge, and electric field change?

Solution — Step by Step

For a parallel plate capacitor with plate area $A$ and separation $d$ :

C_0 = \frac{\varepsilon_0 A}{d}

Without a dielectric, the space between plates contains vacuum (or air).

When a dielectric material is inserted, its molecules (or atoms) become polarised in the presence of the electric field between the plates. The positive end of each dipole aligns toward the negative plate, and vice versa.

This creates an induced surface charge on the dielectric surfaces — opposite in sign to the actual plate charges. This induced charge partially cancels the original field, reducing the net electric field inside the dielectric. The potential difference $V$ across the capacitor would drop if no battery is connected.

But with a dielectric, the capacitance becomes:

C = KC_0 = \frac{K\varepsilon_0 A}{d}

where $K$ is the dielectric constant (also called relative permittivity, $\varepsilon_r$ ). Since $K > 1$ , inserting a dielectric always increases capacitance.

Key condition: Battery remains connected → voltage $V$ is held constant at the battery EMF.

Since $V$ is constant and $C = KC_0 = 3C_0$ :

Quantity	Formula	Change
Capacitance $C$	$C = KC_0$	Increases to $3C_0$ (×K)
Charge $Q$	$Q = CV = KC_0V$	Increases to $3Q_0$ (×K)
Electric field $E$	$E = V/d$	Unchanged (V and d unchanged)
Energy $U$	$U = \frac{1}{2}CV^2 = \frac{1}{2}KC_0V^2$	Increases to $3U_0$ (×K)

When the dielectric is inserted, the capacitor draws more charge from the battery ( $Q$ increases from $Q_0$ to $3Q_0$ ). The battery does work to supply this extra charge. Part of this work increases the stored energy in the capacitor; the rest is dissipated (as heat in circuit resistance).

The battery does work $W_{\text{battery}} = \Delta Q \times V = (3Q_0 - Q_0)V = 2Q_0V$ . The increase in stored energy is $\Delta U = (3U_0 - U_0) = 2U_0 = Q_0V$ . The remaining $Q_0V$ is dissipated.

Why This Works

The dielectric acts as an energy storage enhancer — by polarising, it allows the capacitor to hold more charge for the same voltage. This is why real capacitors always use dielectric materials (ceramic, polyester, mica) rather than just air gaps. Higher dielectric constant → more compact capacitor for the same capacitance.

The “battery connected vs. disconnected” distinction is crucial in JEE problems:

Battery connected: $V$ = constant. $C$ increases → $Q$ increases, $E$ unchanged.
Battery disconnected: $Q$ = constant. $C$ increases → $V$ decreases, $E$ decreases, energy decreases.

Always identify which quantity is being held constant before applying formulas.

Alternative Method

For the disconnected battery case (fixed charge $Q$ ):

$C = KC_0$ → $V = Q/C = Q/(KC_0) = V_0/K$ (voltage drops)
$E = V/d = V_0/(Kd) = E_0/K$ (field drops)
$U = Q^2/(2C) = Q^2/(2KC_0) = U_0/K$ (energy drops — dielectric absorbs the energy)

The energy decrease is the work done by the electric field on the dielectric as it’s pulled in.

Common Mistake

Students often say “dielectric decreases electric field, so it must decrease everything.” This is true only when the battery is disconnected (fixed $Q$ ). When the battery is connected (fixed $V$ ), $E = V/d$ stays constant. The increased charge from the battery compensates for the dielectric’s effect. Always specify the boundary condition — connected or disconnected — before stating what happens to $E$ , $Q$ , or $V$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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