Capacitors

Question 1

Capacitance

Answer 1

The amount of charge an object is able to store per unit potential difference (p.d.) across it
(C = Q / V)

Question 2

Voltage rating of a capacitor

Answer 2

The maximum potential difference that can be safely put across the capacitor

Question 3

When trying to find the capacitance of a capacitor, where do you place the resistor when
a) charging the capacitor?
b) discharging the capacitor?

Answer 3

a) resistor is in parallel with the capacitor
b) resistor is in series with the capacitor

Question 4

An uncharged capacitor is charged with a power supply of voltage V. What is the maximum potential difference across the capacitor?

Answer 4

V

Question 5

Give three uses of capacitors (4 given on this flashcard)

Answer 5

• Camera flashes - charge is dumped into the flash rapidly
• ‘Ultracapacitors’ - provides reliable back-up power for short periods of time
• Smooth out variations in d.c. voltage supplies - capacitor levels out the peaks & troughs
• Burglar alarms - there’s a time delay from when the circuit is closed to when capacitor has discharged enough and the alarm goes off

Question 6

Explain why capacitors are useful for their uses.

Answer 6

They store relatively small charges, but this charge can be stored until its required and discharged rapidly.

Question 7

The energy stored by the capacitor is ____ the energy supplied by the power source.

Answer 7

half

Question 8

Permittivity

Answer 8

A measure of how diffcult it is to generate an electric field in a medium.

Question 9

The higher the permittivity of a material, the ____ charge is needed to generate an electric field of a given size.

Answer 9

more

Question 10

Relative permittivity formula + symbols

Answer 10

ε_r = ε₁ / ε₀
where ε₁ = permittivity of material 1 and ε₀ = permittivity of free space

Question 11

Other name for relative permittivity

Answer 11

dielectric constant

Question 12

What is the dielectric constant?

Answer 12

The ratio of the charge stored with the dielectric between the plates to the charge stored when the dielectric is not present.

Question 13

What is the permittivity of air?

Answer 13

The permittivity of air is only slightly greater than that of free space, so can be assumed to be equal to the permittivity of free space.

Question 14

Describe and explain the rate of charging of a capacitor as charge builds up.

Answer 14

Initially the current through the circuit is high.
Charge builds up on the plates (electrons move towards and build up on the negative plate).
The electrostatic repulsion makes it harder and harder for more electrons to be deposited.
When the p.d. across the capacitor is equal to the p.d. across the supply, the current falls to zero - the capacitor is fully charged.

Question 15

Explain the changes in potential difference and current in a capacitor charging circuit.

Answer 15

• A current starts to flow when the switch is closed.
• The potential difference across the capacitor is zero initially, so there is no p.d. opposing the current.
• The potential difference of the power supply causes an initial relatively high current of V/R to flow (where V is the p.d. of the power supply and R is the resistance of the resistor).
• As the capacitor charges, the p.d. across the capacitor increases, so the p.d. across the resistor decreases, and so the current decreases.

Question 16

What is Q in the formula Q = Q₀(1 - e^-t/RC)

Answer 16

The charge across the capacitor plates at time t (not the charge that has been lost during discharge).

Question 17

What is Q₀ in the formula Q = Q₀(1 - e^-t/RC)

Answer 17

The charge of the capacitor plates when fully charged (not the charge that has been lost during discharge).

Question 18

When the discharge time t is equal to RC, the equation for the charge left on a discharging capacitor becomes what?
Therefore what is Q / Q₀ equal to?

Answer 18

Q = Q₀e^-1
therefore Q / Q₀ = 1/e = 1 / 2.718…
~ 0.37

Question 19

What is the time constant definition, in terms of charge, for a capacitor that is
a) charging?
b) discharging?

Answer 19

a) Time taken for a capacitor to charge up to 63% of its maximum charge.
b) Time taken for a capacitor to discharge down to 37% of its maximum charge.

Question 20

What is the time constant definition, in terms of potential difference, for a capacitor that is
a) charging?
b) discharging?
Explain your answer.

Answer 20

a) Time taken for the potential difference across a capacitor to increase to 63% of the source potential difference.
b) Time taken for the potential difference across a capacitor to decrease to 37% of the source potential difference.

Question 21

The larger the resistance in series with the capacitor, the ____ it takes to charge or discharge. Explain your answer.

Answer 21

longer
A greater proportion of the input potential difference is dropped across the resistor so a lower proportion is dropped across the capacitor.

Question 22

In practice, the time taken for a capacitor to
charge or discharge fully is taken to be about aRC or b𝜏. What are the values of a and b?

Answer 22

a = b = 5

Question 23

Time to halve definitions

Involves multiple values

Answer 23

The time taken for the charge, current or potential difference of a discharging capacitor to decrease to half of the initial value.

Question 24

formula for time for the charge across a capacitor to halve

Not given in the formula booklet

Answer 24

T_1/2 = 0.69RC

Question 25

Derive the formula for time for the charge across a capacitor to halve.

Answer 25

* Time to halve is when Q = 1/2 Q₀ * Therefore, for a discharging capacitor, Q = Q₀e^{-T_1/2/RC * This becomes 1/2 Q₀ = Q₀e-T_1/2/RC * => 1/2 = e-T_1/2/RC * ln-ing both sides: ln(1/2) = ln(e-T_1/2/RC) * ln(1/2) = -T_1/2/RC * T_1/2 = - ln(1/2) * RC = 0.693...RC * T_1/2 ~ 0.69RC}

Question 26

How many significant figures should a result using the time to halve formula be given to? Explain your answer.

Answer 26

The 0.69 in T_1/2 = 0.69RC comes from ln(2) = 0.6931 which is rounded to (and so accurate to) 2 s.f. only. This means answers to questions that use this formula shouldn’t be given to more than 2 s.f.

Question 27

A capacitor of capacitance C has a charge of Q stored on the plates. The potential difference between the plates is doubled. What is the change in the energy stored by the capacitor?

Answer 27

ΔE = E2 - E1 = 1/2 CV² - 1/2 Cv² = 1/2 C(2V)² - 1/2 Cv² = 2CV² - 1/2 Cv² = 3/2 CV² = 3/2 C (Q/C)² = 3/2 Q² / C

Question 28

Explain how a dielectric changes the capacitance of a capacitor.

Answer 28

* When charge is applied to the plates of a capacitor, an electric field is generated between them (positive to negative). * The negative ends of the polar molecules in the dielectric are attracted to the positively charged plate, and vice versa. * This causes the molecules to rotate to align themselves **anti-parallel** to the electric field between the plates. * The polar molecules produce their own electric field, but since the charges are opposite to that of the capacitor plates, the **molecules' field acts in the opposite** direction. * This **reduces the overall electric field** between the capacitor plates. * Since E = V / d, the **p.d. between the plates reduces**, but as the **charge remains constant**, the capacitor's capacitance must increase (C = Q / V).

Question 29

State what is meant by a dielectric constant of 5.

Answer 29

Ratio of permittivity of a medium to the permittivity of free space permittivity of a medium / permittivity of free space or ε_r = ε₁ / ε₀ where ε₁ = permittivity of material 1 and ε₀ = permittivity of free space

Question 30

Total capacitance of capacitors in parallel

Answer 30

C total of capacitors in parallel = C1 + C2 + C3 + ...

Question 31

Total capacitance of capacitors in series

Answer 31

1 / C total of capacitors in series = (1 / C1) + (1 / C2) + (1 / C3) + ...

Question 32

Initially, a charged capacitor stores 1600 microjoules of energy. When the p.d. across it decreases by 2.0V, the energy stored by it becomes 400 microjoules. What is the capacitance of this capacitor?

Answer 32

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Question 33

The switch S in the circuit is held in position 1, so that the capacitor C becomes fully charged to a pd V and stores energy E. The switch is then moved quickly to position 2, allowing C to discharge through the fixed resistor R. It takes 36ms for the pd across C to fall to V/2. **After the switch has been moved to position 2**, how long does it take before the energy stored by C has fallen to E/16? A) 51ms B) 72ms C)432ms D) 576ms

Answer 33

https://www.thestudentroom.co.uk/showthread.php?t=2157867&page=37 B - 72 ms. Note: for the second use of the decay formula, V₀ = 1 since we are timing from when the switch moved and not the time from when V = V / 2. | Reply 732