Worked examples · Sheet 2

Parallel-plate capacitors

Theory used: Capacitor Parallel-plate capacitor Capacitance of an isolated conductor How to compute a capacitance

Exercise 1

A parallel-plate capacitor has square plates of side $ℓ = 20 cm$ at a distance $d = 1.0 mm$ , in vacuum, and is connected to a source that maintains the potential difference $Δ V = 50 V$ . Compute the capacitance, the charge on the plates, the surface charge density and the electric field between the plates.

C = \frac{ε _{0} A}{d}, A = ℓ^{2}, Q = C Δ V, σ = \frac{Q}{A}, E = \frac{Δ V}{d}

C Q σ E = \frac{8.85 \cdot 1 0 ^{- 12} \cdot 0.040}{1.0 \cdot 1 0 ^{- 3}} F \approx 3.5 \cdot 1 0^{- 10} F = 354 pF = 3.54 \cdot 1 0^{- 10} \cdot 50 C \approx 1.8 \cdot 1 0^{- 8} C = 18 nC = \frac{1.77 \cdot 1 0 ^{- 8}}{0.040} C/ m^{2} \approx 4.4 \cdot 1 0^{- 7} C/ m^{2} = \frac{50}{1.0 \cdot 1 0 ^{- 3}} V/m = 5.0 \cdot 1 0^{4} V/m

The distance between the plates is much smaller than the side, therefore edge effects are negligible and the formula of the parallel-plate capacitor applies, with the area $A = 0.2 0^{2} m^{2} = 0.040 m^{2}$ . The source fixes the potential difference, hence the charge follows from the definition of capacitance, $Q = C Δ V$ . The density is the charge divided by the area, because the charges are distributed uniformly on the facing sides.

The field between the plates is uniform, therefore it is the ratio of the potential difference to the distance. As a check, Coulomb’s theorem gives $E = \frac{σ}{ε _{0}} = \frac{4.43 \cdot 1 0 ^{- 7}}{8.85 \cdot 1 0 ^{- 12}} V/m = 5.0 \cdot 1 0^{4} V/m$ , the same value.

Exercise 2

We want to build a parallel-plate capacitor in vacuum with capacitance $C = 1.0 μ F$ , with the plates at a distance $d = 0.10 mm$ . Compute the area of the plates and the side of square plates. Also compute the charge and the field between the plates when the capacitor is connected to $12 V$ .

C = \frac{ε _{0} A}{d} ⟹ A = \frac{C d}{ε _{0}}, ℓ = A

A ℓ Q E = \frac{1.0 \cdot 1 0 ^{- 6} \cdot 1.0 \cdot 1 0 ^{- 4}}{8.85 \cdot 1 0 ^{- 12}} m^{2} \approx 11 m^{2} = 11.3 m \approx 3.4 m = C Δ V = 1.0 \cdot 1 0^{- 6} \cdot 12 C = 12 μ C = \frac{Δ V}{d} = \frac{12}{1.0 \cdot 1 0 ^{- 4}} V/m = 1.2 \cdot 1 0^{5} V/m

The area is obtained from the formula of the parallel-plate capacitor, since capacitance and distance are given. About $11 m^{2}$ is needed, that is, two square sheets more than three meters on a side held a tenth of a millimeter apart. The result shows that the microfarad is a large capacitance for a capacitor in vacuum; in real capacitors the plates are thin rolled-up strips separated by a dielectric, which multiplies the capacitance by $ε_{r}$ .

The charge and the field follow from the definition of capacitance and from the uniformity of the field. The field, $1.2 \cdot 1 0^{5} V/m$ , is smaller than the dielectric strength of air, about $3 \cdot 1 0^{6} V/m$ , therefore the voltage of $12 V$ can be applied even with air between the plates.

Exercise 3

A parallel-plate capacitor with plates of area $A = 0.020 m^{2}$ at a distance $d_{1} = 2.0 mm$ is charged by connecting it to a $Δ V_{1} = 100 V$ source. The plates are then moved to the distance $d_{2} = 5.0 mm$ . Compute capacitance, charge, potential difference and field after the displacement in two cases: (a) the capacitor is disconnected from the source before the displacement; (b) the capacitor remains connected to the source.

C_{1} = \frac{ε _{0} A}{d _{1}}, C_{2} = \frac{ε _{0} A}{d _{2}} = \frac{d _{1}}{d _{2}} C_{1}, Q_{1} = C_{1} Δ V_{1}

(a) Q_{2} = Q_{1}, Δ V_{2} = \frac{Q _{1}}{C _{2}} = \frac{d _{2}}{d _{1}} Δ V_{1}, E_{2} = \frac{Δ V _{2}}{d _{2}} = \frac{Δ V _{1}}{d _{1}}

(b) Δ V_{2} = Δ V_{1}, Q_{2} = C_{2} Δ V_{1} = \frac{d _{1}}{d _{2}} Q_{1}, E_{2} = \frac{Δ V _{1}}{d _{2}}

C_{1} Q_{1} (a) Δ V_{2} (b) Q_{2} = \frac{8.85 \cdot 1 0 ^{- 12} \cdot 0.020}{2.0 \cdot 1 0 ^{- 3}} F = 88.5 pF, C_{2} = \frac{2.0}{5.0} \cdot 88.5 pF = 35.4 pF = 88.5 \cdot 1 0^{- 12} \cdot 100 C \approx 8.9 nC = \frac{5.0}{2.0} \cdot 100 V = 250 V, E_{2} = \frac{100}{2.0 \cdot 1 0 ^{- 3}} V/m = 5.0 \cdot 1 0^{4} V/m = \frac{2.0}{5.0} \cdot 8.85 nC \approx 3.5 nC, E_{2} = \frac{100}{5.0 \cdot 1 0 ^{- 3}} V/m = 2.0 \cdot 1 0^{4} V/m

In both cases the capacitance depends only on the geometry, therefore it goes from $88.5 pF$ to $35.4 pF$ , that is, it is reduced by the ratio $\frac{d _{1}}{d _{2}}$ . The problem is solved by identifying the quantity that stays constant.

In case (a) the capacitor is isolated, hence the charge is conserved. The potential difference $\frac{Q}{C}$ grows by the same ratio by which the capacitance decreases, and becomes $250 V$ . The field does not change, because it depends only on the charge density, $E = \frac{σ}{ε _{0}}$ , which stays the same.

In case (b) the source maintains $100 V$ , therefore the charge drops to $C_{2} Δ V_{1} \approx 3.5 nC$ ; the source takes back the excess charge. The field, equal to $\frac{Δ V}{d}$ , decreases to $2.0 \cdot 1 0^{4} V/m$ .