Conductors, Dielectrics and Capacitors: Quick Revision

A brief note on Conductors, Dielectrics and Capacitors. Read it before a test. Try each question at the end before you open its answer.

Conductors at rest

• Inside: the field is zero, there is no charge, and the whole conductor has one potential.
• Surface: the extra charge sits there. The field just outside is at right angles to it, $E = \dfrac{\sigma}{\varepsilon_0}$.
• Shielding: a closed hollow conductor keeps the field of outside charges out of its cavity. A charge $+q$ inside draws $-q$ on the inner wall and $+q$ on the outer wall.
• Joined spheres: the same potential, so $\dfrac{q_1}{q_2} = \dfrac{R_1}{R_2}$ and $\dfrac{\sigma_1}{\sigma_2} = \dfrac{R_2}{R_1}$. The charge crowds on sharp points.

Dielectrics

• Polarisation: polar molecules line up, and non-polar molecules get an induced dipole. Bound charge appears on the faces.
• Dielectric constant: $K = \dfrac{E_0}{E}$, so $E = \dfrac{E_0}{K}$. A metal acts like $K = \infty$.
• Bound charge: $\sigma_b = \sigma\left(1 - \dfrac{1}{K}\right)$.

Capacitance

• Definition: $C = \dfrac{Q}{V}$, in farad. It depends on the shape, the size and the material, not on $Q$ or $V$.
• Isolated sphere: $C = 4\pi\varepsilon_0 R$.
• Parallel plates: $C = \dfrac{\varepsilon_0 A}{d}$, with $E = \dfrac{V}{d}$.
• Concentric spheres: $C = 4\pi\varepsilon_0\dfrac{ab}{b - a}$.
• Breakdown: air sparks at about $3 \times 10^{6}$ V/m, so $V_{\max} = E_{\max}d$.

Dielectric slabs

• Fills the gap: $C = KC_0$.
• Battery connected: $V$ and $E$ stay, and $Q$ grows by $K$.
• Battery removed: $Q$ stays, and $V$ and $E$ fall by $K$.
• Slab of thickness $t$: $C = \dfrac{\varepsilon_0 A}{d - t + t/K}$, wherever the slab sits.
• Metal slab: $C = \dfrac{\varepsilon_0 A}{d - t}$.

A worked example

Plates of area $250\ \text{cm}^2$ are $1.2$ mm apart in air. A $15$ V battery is joined to them. Find $C$ and $Q$.

1. Capacitance. $C = \dfrac{8.85 \times 10^{-12} \times 250 \times 10^{-4}}{1.2 \times 10^{-3}} = 1.84 \times 10^{-10}\ \text{F} = 184\ \text{pF}$.
2. Charge. $Q = CV = 184\ \text{pF} \times 15\ \text{V} = 2.77\ \text{nC}$.

Practise with new numbers

Check yourself

1. A capacitor charged to $24$ V is taken off its battery. A slab of dielectric constant $4$ fills the gap. What is the voltage now?

2. Two metal spheres of radius $R$ and $2R$ are joined by a wire. Which one has the larger charge on each unit of area?

3. Why can a dielectric not make the field between plates zero?