Formulas/physics/Electrostatic Potential/Capacitance with Dielectric

Capacitance with Dielectric

Inserting a dielectric of dielectric constant K between the plates multiplies capacitance by K. K > 1 always. K = 1 for vacuum.

Class 11Class 12

Derivation

What a dielectric does

When a dielectric is inserted between capacitor plates, the electric field polarises the molecules — positive charges shift slightly in the field direction, negative charges against it. This creates a surface charge on the dielectric faces that partially cancels the field from the plate charges.

Field reduction

Let $E_0 = \sigma/\varepsilon_0$ be the original field (no dielectric). The polarisation creates an opposing field $E_p$ . Net field:

E = E_0 - E_p = \frac{E_0}{K}

where $K > 1$ is the dielectric constant (relative permittivity) of the material.

New potential difference and capacitance

V = Ed = \frac{E_0 d}{K} = \frac{V_0}{K}

Since charge $Q$ is unchanged (isolated capacitor):

C = \frac{Q}{V} = \frac{Q}{V_0/K} = K\cdot\frac{Q}{V_0} = KC_0

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

Key facts

$K \geq 1$ always; $K = 1$ for vacuum
For a capacitor connected to a battery (constant $V$ ): inserting dielectric increases $Q$ and $C$ , energy increases
For an isolated capacitor (constant $Q$ ): inserting dielectric decreases $V$ and $E$ , energy decreases by factor $K$

Remember

The dielectric constant $K$ is also written as $\varepsilon_r$ (relative permittivity). The absolute permittivity of the medium is $\varepsilon = K\varepsilon_0 = \varepsilon_r\varepsilon_0$.

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