Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Worked example 3.4.1

Two short calculations tied to the applications. Each uses only $D = ε E$ , the boundary conditions, $C = Q / V$ and $W = \frac{1}{2} \int D \cdot E d v$ . No new physics is needed. Try each one yourself first: write down the formula from the corresponding derivation (Problem 3.4.1, the energy of a capacitor) and only then insert the numbers.

Worked example 3.4.1

Partly filled plate capacitor

cf. Application 3.4.1

Given. Plates of area $S = 100$ cm $^{2}$ , $d = 2$ mm apart, in air. A mica slab ( $ε_{r} = 6$ ), $t = 1$ mm thick, lies on the lower plate, and $V = 1$ kV.

Find. (a) $C$ , compared with the empty and the completely filled gap, (b) $E$ in the air and in the mica, (c) $ρ_{S}$ on the plates and the bound charge on the mica, and (d) the highest voltage before breakdown.

Parallel plates 2 mm apart with a 1 mm mica slab on the lower plate and air above

Try it yourself first.

Strategy

The mica face is parallel to the plates and carries no free charge, so the normal component of $D$ is continuous there: $D$ is the same in the air and in the mica, while $E = D / ε$ jumps by the factor $ε_{r}$ . The voltage is the sum of the two uniform-field contributions, field times thickness.

(a)

With $D_{air} = D_{mica} = D$ ,

E_{air} = \frac{D}{ε _{0}}, E_{mica} = \frac{D}{ε _{r} ε _{0}},

V = E_{air} (d - t) + E_{mica} t = \frac{D}{ε _{0}} [(d - t) + \frac{t}{ε _{r}}] .

The plate charge is $Q = ρ_{S} S = D S$ , hence

C = \frac{Q}{V} = \frac{ε _{0} S}{( d - t ) + t / ε _{r}} = \frac{8.842 \times 1 0 ^{- 12} \times 1 0 ^{- 2}}{( 1 + 0.1667 ) \times 1 0 ^{- 3}} = 75.8 pF .

For comparison, $C_{0} = ε_{0} S / d = 44.2$ pF with the gap empty and $6 C_{0} = 265$ pF with the gap completely filled: half a gap of mica raises $C$ only by a factor $1.7$ , not $6$ .

(b)

$D = ρ_{S} = C V / S = 7.58 \times 1 0^{- 6}$ C/m $^{2}$ , so

E_{air} E_{mica} = \frac{D}{ε _{0}} = 857 kV/m, = \frac{D}{6 ε _{0}} = 143 kV/m .

Check: $857 \times 1 0^{3} \times 1 0^{- 3} + 143 \times 1 0^{3} \times 1 0^{- 3} = 1000$ V. $✓$

(c)

$ρ_{S} = \pm 7.58$ μC/m $^{2}$ on the plates, i.e. $Q = 75.8$ nC. In the mica

P = (ε_{r} - 1) ε_{0} E_{mica} = 5 \times 8.842 \times 1 0^{- 12} \times 1.43 \times 1 0^{5} = 6.32 μ C/ m^{2},

so the mica faces carry the bound charges $\mp 6.32$ μC/m $^{2}$ (opposite in sign to the free charge on the plate side they face): five sixths of the plate charge is cancelled inside the mica.

(d)

The field in the air is six times larger, so the air breaks down first. With $E_{air} = 3$ MV/m, $E_{mica} = 0.5$ MV/m and

V_{m a x} = 3 \times 1 0^{6} \times 1 0^{- 3} + 0.5 \times 1 0^{6} \times 1 0^{- 3} = 3.5 kV,

less than the $3 \times 1 0^{6} \times 2 \times 1 0^{- 3} = 6$ kV of the empty $2$ mm gap.

Answer

C = 75.8 pF, E_{air} = 857 kV/m, E_{mica} = 143 kV/m, V_{m a x} = 3.5 kV

Worked example 3.4.1, with x measured up from the lower plate. Left: the field steps from 143 kV/m in the mica to 857 kV/m in the air. Right: the mica takes only 143 V of the 1000 V, and the air layer takes the rest (dashed: the linear potential of the empty gap). — **Figure 3.4.36.** Worked example 3.4.1, with $x$ measured up from the lower plate. Left: the field steps from $143$ kV/m in the mica to $857$ kV/m in the air. Right: the mica takes only $143$ V of the $1000$ V, and the air layer takes the rest (dashed: the linear potential of the empty gap).

Potential across the partly filled gap: 143 V across the mica and the rest across the air

Check

$C$ lies between the empty ( $44.2$ pF) and the filled ( $265$ pF) values, and the plate charge $Q = C V = 75.8$ nC agrees with $D S$ . $✓$

What the numbers say

The capacitor is two capacitors in series (Home exercise 3.4.1): $1/ C = (d - t) / (ε_{0} S) + t / (ε_{r} ε_{0} S)$ . The slab takes only $1/7$ of the voltage. The air gap takes the rest, has the higher field and breaks down first (Figure 3.4.36). A dielectric that does not fill the gap completely therefore lowers the voltage rating. The same effect makes a gas gap inside a solid insulation dangerous: by the continuity of $D_{n}$ the gas sees $ε_{r}$ times the field of the solid, while it breaks down at far lower fields.

Worked example 3.4.1: Partly filled plate capacitor