Fill the gap of a capacitor with an insulator such as glass, plastic or oil, and its capacitance goes up by a factor $\kappa$, the material's dielectric constant. What else changes depends on one question: is the battery still connected?
Key idea
A dielectric that fills the gap gives $C = \kappa C_0$.
Battery connected (fixed $V$): $Q \to \kappa Q_0$ and $U = \tfrac{1}{2}CV^2 \to \kappa U_0$.
Isolated (fixed $Q$): $V \to V_0/\kappa$, $E \to E_0/\kappa$ and $U = Q^2/(2C) \to U_0/\kappa$.
Battery connected (fixed $V$): $Q \to \kappa Q_0$ and $U = \tfrac{1}{2}CV^2 \to \kappa U_0$.
Isolated (fixed $Q$): $V \to V_0/\kappa$, $E \to E_0/\kappa$ and $U = Q^2/(2C) \to U_0/\kappa$.
Why it works
An insulator has no free electrons, but its molecules can stretch or turn. In the field between the plates they line up a little, positive ends toward the negative plate. Inside the material these tiny dipoles cancel each other, but at the two faces a thin layer of bound charge is left over, opposite in sign to the nearby plate. That bound charge cancels part of the plates' field. With a weaker field for the same free charge, the voltage is smaller, and $C = Q/V$ is larger.
| Battery connected | Isolated | |
|---|---|---|
| $C$ | $\kappa C_0$ | $\kappa C_0$ |
| $V$ | same | $V_0/\kappa$ |
| $Q$ | $\kappa Q_0$ | same |
| $E$ between plates | same ($V/d$) | $E_0/\kappa$ |
| $U$ | $\kappa U_0$ | $U_0/\kappa$ |
Worked examples
Worked example: Isolated
A $50\ \text{pF}$ air capacitor is charged to $12\ \text{V}$, disconnected, and filled with a dielectric of $\kappa = 3.0$. Find the new voltage and energy.
- $Q = C_0V_0 = 600\ \text{pC}$ is trapped.
- $C = 3.0 \times 50 = 150\ \text{pF}$, so $V = Q/C = 600/150 = 4.0\ \text{V}$.
- $U = Q^2/(2C)$ drops from $3.6\ \text{nJ}$ to $1.2\ \text{nJ}$. The difference went into pulling the slab in: a capacitor draws a dielectric toward it.
Answer: $4.0\ \text{V}$, $1.2\ \text{nJ}$
Worked example: Connected
Same capacitor, but it stays on the $12\ \text{V}$ battery while the dielectric goes in.
- $V$ stays $12\ \text{V}$; $C$ becomes $150\ \text{pF}$.
- $Q = CV = 1800\ \text{pC}$: the battery pushed $1200\ \text{pC}$ more onto the plates.
- $U = \tfrac{1}{2}CV^2 = 10.8\ \text{nJ}$, three times as much. The battery supplied the extra energy (and more, some of which pulled the slab in).
Answer: $1800\ \text{pC}$, $10.8\ \text{nJ}$
Worked example: Half filled
A dielectric of $\kappa = 3.0$ fills only half of a $50\ \text{pF}$ air capacitor. Find $C$ if it fills (a) the left half of the area, full gap, (b) half the thickness, full area.
- (a) Side by side, both halves feel the same voltage: two capacitors in parallel, $\dfrac{C_0}{2} + \dfrac{\kappa C_0}{2} = 25 + 75 = 100\ \text{pF}$.
- (b) Stacked, the same charge passes through both layers: two capacitors in series, each with half the gap ($2C_0$ and $2\kappa C_0$): $\dfrac{1}{C} = \dfrac{1}{100} + \dfrac{1}{300}$, so $C = 75\ \text{pF}$.
Answer: (a) $100\ \text{pF}$, (b) $75\ \text{pF}$
Common mistake
Not tracking what stays fixed. "The dielectric raises $C$, so $Q = CV$ goes up" is right only with the battery connected. Unplugged, the charge has nowhere to come from, so $Q$ stays put and $V$ falls instead. Before any dielectric problem, write down which is fixed: $V$ (connected) or $Q$ (isolated). Then use the energy formula built on that fixed quantity: $\tfrac{1}{2}CV^2$ for fixed $V$, $Q^2/(2C)$ for fixed $Q$.