Questions on capacitors and the dielectric

Question 1

Define capacitance and state its unit.

Accepted Answer

Charge stored per unit potential difference, C = Q/V (1); unit: farad (F), equal to one coulomb per volt (1).

Question 2

A capacitor stores a charge of 6.0 × 10⁻³ C when the potential difference across it is 12 V. Calculate its capacitance.

Accepted Answer

C = Q/V = (6.0 × 10⁻³)/12 (1) C = 5.0 × 10⁻⁴ F (500 μF) (1)

Question 3

Write down three equivalent expressions for the energy stored by a charged capacitor.

Accepted Answer

E = ½QV (1); E = ½CV² (1); E = ½Q²/C (1).

Question 4

A capacitor is charged in stages and a graph of charge stored against potential difference is plotted. State what is represented by the gradient of this graph.

Accepted Answer

The capacitance of the capacitor, since the gradient is Q/V = C (1).

Question 5

The capacitor in a defibrillator stores a charge of 0.36 C when charged to a potential difference of 1.8 kV. Calculate its capacitance.

Accepted Answer

C = Q/V = 0.36/1800 (1) C = 2.0 × 10⁻⁴ F (200 μF) (1)

Question 6

A 50 μF capacitor is used in a signal filter. Calculate the potential difference needed for it to store a charge of 4.0 mC.

Accepted Answer

V = Q/C = (4.0 × 10⁻³)/(50 × 10⁻⁶) (1) V = 80 V (1)

Question 7

A 100 μF capacitor is charged to a potential difference of 20 V. Calculate the charge stored and the energy stored.

Accepted Answer

Q = CV = 100 × 10⁻⁶ × 20 = 2.0 × 10⁻³ C (1) E = ½CV² = ½ × 100 × 10⁻⁶ × 20² (1) E = 0.020 J (1)

Question 8

A capacitor stores a charge of 5.0 × 10⁻³ C at a potential difference of 8.0 V. Calculate the energy stored.

Accepted Answer

E = ½QV = ½ × 5.0 × 10⁻³ × 8.0 (1) E = 0.020 J (1)

Question 9

A parallel-plate capacitor has plates of area 0.020 m² separated by a 1.0 mm vacuum gap. Calculate its capacitance. ε₀ = 8.85 × 10⁻¹² F m⁻¹

Accepted Answer

C = ε₀A/d = (8.85 × 10⁻¹² × 0.020)/(1.0 × 10⁻³) (1) C = 1.77 × 10⁻¹⁰ F (177 pF) (1)

Question 10

A 470 μF capacitor in a camera flash unit is charged to 9.0 V. Calculate the charge stored and the energy stored.

Accepted Answer

Q = CV = 470 × 10⁻⁶ × 9.0 = 4.23 × 10⁻³ C (1) E = ½CV² = ½ × 470 × 10⁻⁶ × 9.0² (1) E = 0.019 J (1)

Question 11

The sensing pad of a touch screen behaves as a parallel-plate capacitor. The pad has an area of 4.0 cm² and the plates are separated by a 0.10 mm layer of glass with relative permittivity 5.0. Calculate the capacitance of the pad.
ε₀ = 8.85 × 10⁻¹² F m⁻¹

Accepted Answer

A = 4.0 × 10⁻⁴ m² and d = 1.0 × 10⁻⁴ m (1) C = ε₀ε_rA/d = (8.85 × 10⁻¹² × 5.0 × 4.0 × 10⁻⁴)/(1.0 × 10⁻⁴) (1) C = 1.77 × 10⁻¹⁰ F (177 pF) (1)

Question 12

A student charges a capacitor from a variable supply and plots the potential difference across it against the charge stored. The graph is a straight line through the origin that passes through Q = 3.6 mC at V = 12 V. Determine the capacitance of the capacitor, and use the graph to determine the energy stored at 12 V.

Accepted Answer

The gradient is V/Q = 1/C, so C = Q/V = (3.6 × 10⁻³)/12 (1) C = 3.0 × 10⁻⁴ F (300 μF) (1) Energy = area under the line = ½QV = ½ × 3.6 × 10⁻³ × 12 (1) E = 0.022 J (1)

Question 13

A 1500 μF capacitor in a bicycle rear light is charged to 6.0 V. Show that the energy stored is about 27 mJ. Go on to calculate the average power delivered when the capacitor discharges completely through the LED in 0.30 s.

Accepted Answer

E = ½CV² = ½ × 1500 × 10⁻⁶ × 6.0² (1) E = 0.027 J = 27 mJ (1) P = E/t = 0.027/0.30 = 0.090 W (1)

Question 14

The capacitor in the earlier question (177 pF with a vacuum gap) has the gap filled with a dielectric of relative permittivity 4.0. Calculate the new capacitance and state the effect on the energy stored if the capacitor remains connected to the same supply voltage.

Accepted Answer

C = ε₀ε_rA/d, so C increases by a factor of 4.0 (1) C = 4.0 × 177 = 708 pF (1) At fixed V, E = ½CV², so the energy stored also increases by a factor of 4.0 (1).

Question 15

A 220 μF capacitor stores a charge of 3.3 × 10⁻³ C. Calculate the energy stored and the potential difference across the capacitor.

Accepted Answer

E = ½Q²/C = ½ × (3.3 × 10⁻³)²/(220 × 10⁻⁶) (1) E = 0.025 J (1) V = Q/C = (3.3 × 10⁻³)/(220 × 10⁻⁶) (1) V = 15 V (1)

Question 16

Explain how inserting a polar dielectric between the plates of a capacitor increases its capacitance.

Accepted Answer

The polar molecules rotate to align with the electric field between the plates (1); the aligned molecules produce a field that opposes the applied field, reducing the net field and hence the potential difference for a given charge (1); since C = Q/V, more charge is stored per volt, so the capacitance increases (1).

Question 17

A designer needs a capacitor for a memory back-up circuit. It must store at least 2.5 J when charged to its maximum rated potential difference. Two capacitors are available:
capacitor P: capacitance 0.10 F, maximum rated pd 6.3 V
capacitor Q: capacitance 0.022 F, maximum rated pd 16 V
Deduce which capacitor, if either, meets the requirement.

Accepted Answer

Use of E = ½CV² at the maximum rated pd (1)
E_P = ½ × 0.10 × 6.3² = 1.98 J (1)
E_Q = ½ × 0.022 × 16² = 2.82 J (1)
Only Q stores more than 2.5 J; P falls short, so the designer should choose Q (1).

Question 18

A parallel-plate capacitor containing a polymer dielectric sheet of relative permittivity 2.5 is charged and then disconnected from the supply. The sheet is then withdrawn from between the plates. State and explain what happens to the capacitance, the charge stored, the potential difference across the plates and the energy stored.

Accepted Answer

C = ε₀ε_rA/d, so removing the dielectric reduces the capacitance by a factor of 2.5 (1). The charge is unchanged: the capacitor is disconnected, so there is no path for charge to leave the plates (1). V = Q/C, so the pd increases by a factor of 2.5 (1). E = ½Q²/C increases by a factor of 2.5; the extra energy is the work done against the attraction of the plates in withdrawing the sheet (1).

Question 19

A camera flash unit uses a capacitor, charged by a small battery, to supply a xenon flash tube. Explain why the energy stored by the charged capacitor is ½QV rather than QV, and explain why a capacitor, rather than the battery alone, is used to supply the flash tube.

Accepted Answer

The first charge transferred arrives when the capacitor pd is (almost) zero, so it needs (almost) no work (1); as charge builds up the pd rises (in proportion, since V = Q/C), so each successive amount of charge is moved through a larger pd (1); the average pd during charging is V/2, so the energy stored is Q × V/2 = ½QV, the area under the graph of pd against charge (1). The capacitor can release its whole store in a very short time (about a millisecond) (1); this gives a very large power, P = E/t, far greater than the battery could deliver directly (1); the battery instead recharges the capacitor slowly at low power between flashes (1).

Question 20

Two capacitors, of capacitance C₁ and C₂, are first connected in parallel across a supply of pd V, and then reconnected in series across the same supply. For the parallel connection, starting from Q = CV and the fact that both capacitors are connected across the same pd, show that the combined capacitance is C = C₁ + C₂. For the series connection, starting from the fact that the two capacitors carry the same charge Q, show that the combined capacitance obeys 1/C = 1/C₁ + 1/C₂.

Accepted Answer

Parallel: both capacitors have the same pd V across them, so Q₁ = C₁V and Q₂ = C₂V (1); the total charge supplied is their sum, Q = Q₁ + Q₂ = (C₁ + C₂)V (1); dividing by V and using C = Q/V gives C = C₁ + C₂ (1). Series: the conductor between the two capacitors is isolated and was initially neutral, so it must stay neutral, which forces both capacitors to carry the same charge Q (1); each pd is V₁ = Q/C₁ and V₂ = Q/C₂, and the pds add, V = V₁ + V₂ (1); substituting and dividing through by Q gives 1/C = 1/C₁ + 1/C₂ (1).

Question 21

A 4.7 μF capacitor and a 10 μF capacitor are connected in series across a 12 V supply. Calculate the combined capacitance, the charge stored and the potential difference across each capacitor. The two capacitors are then reconnected in parallel across the same 12 V supply; calculate the energy stored by the combination in that arrangement.

Accepted Answer

1/C = 1/C₁ + 1/C₂ = 1/4.7 + 1/10, in reciprocal microfarads (1)
C = 3.2 μF, below the smaller of the two, as a series combination must be (1)
Q = CV = 3.2 × 10⁻⁶ × 12 = 38 μC, carried by both capacitors alike, because the conductor between them is isolated and stays neutral (1)
V = Q/C for each: 8.2 V across the 4.7 μF and 3.8 V across the 10 μF, which add to the 12 V supply (1)
In parallel C = C₁ + C₂ = 14.7 μF (1)
E = ½CV² = ½ × 14.7 × 10⁻⁶ × 12² = 1.1 mJ (1)

Question 22

A 0.50 F supercapacitor is charged to 5.0 V and used to keep a memory chip alive during a power cut. The chip stops working once the potential difference across the capacitor has fallen to 3.0 V. By considering the area under the graph of potential difference against charge, determine the energy the capacitor can deliver to the chip, and state how much of its full store that leaves unused.

Accepted Answer

Q = CV at each pd: 2.5 C at 5.0 V and 1.5 C at 3.0 V, so the charge delivered is 1.0 C (1)
The energy delivered is the area under the line between those two charges, a trapezium with parallel sides 5.0 V and 3.0 V, so the charge is delivered at a mean pd of 4.0 V (1)
E = ½(5.0 + 3.0) × 1.0 (1)
E = 4.0 J (1)
The same result follows as the difference of two triangles, ½CV² at each pd: 6.25 − 2.25 = 4.0 J. The 2.25 J still held below 3.0 V cannot be used, so the chip receives only 4.0 J of the 6.25 J stored (1)

Question 23

A 4700 μF capacitor is charged to 12 V and then disconnected from the supply. Its two terminals are then joined by short leads to the terminals of a second, identical 4700 μF capacitor, which is uncharged. Calculate the potential difference across the two capacitors once the charge has settled, and determine the energy stored before and after the connection. Explain what has become of the difference.

Accepted Answer

Q = CV = 4700 × 10⁻⁶ × 12 = 0.0564 C, and charge is conserved: the two capacitors hold that same total afterwards (1)
They are identical and end up with the same pd across both, so each holds 0.0282 C, and V = Q/C gives 6.0 V, half the original pd (1)
Before: E = ½CV² = ½ × 4700 × 10⁻⁶ × 12² = 0.338 J (1)
After: two stores of ½CV² at 6.0 V, a total of 0.169 J; the same follows from ½QV with the whole 0.0564 C now sitting at 6.0 V (1)
0.169 J, half the original store, has gone. The charge that moved passed through the resistance of the leads and the capacitors themselves, warming them, and none of it is recoverable (1)