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Physics > Y2: Capacitors > Flashcards

Y2: Capacitors Flashcards

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1
Q

What is a capacitor

A

An electrical component that can store an electric charge. It is made of 2 electrical conducting plates separated by an electrical insulator (Dielectric).

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2
Q

What is the voltage rating of a capacitor

A

The maximum potential difference that can be safely put across the capacitor.
If the power supply has a V lower than this, it will charge to a maximum value equal to the pd of the battery.

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3
Q

What is capacitance (C)

A

The amount of charge a capacitor is able to store per unit potential difference across it
(Measured in Farads: 1F = 1CV^1)

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4
Q

What is the equation for capacitance (relating to charge and potential difference)

A

C = Q/V

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5
Q

How would you determine the capacitance of a capacitor with current and potential difference

A
  • Connect the capacitor in series with a battery, variable resistor and ammeter, and in parallel with a voltmeter
  • Start a timer and constantly adjust the variable resistor to keep the current constant
  • Record the pd at regular intervals
  • Calculate Q for these times as Q=It
  • Plot a Q-V graph to show a directly proportional relationship
  • C = Q/V, so the capacitance is given by the gradient of the graph
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6
Q

What are some possible uses for capacitors

A
  • Camera flash:
    Battery charges capacitor over a few seconds, then entire charge is released to give a short, bright flash
  • Ultracapacitors: (Really big capacitors…)
    Used as reliable backup power supplies for short periods
  • Smooths out dc voltage supply:
    Power from capacitor absorbs peaks and fills troughs in a fluctuating power supply
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7
Q

Why is energy stored by capacitors

A

The oppositely charged plates repel, but are held together by electrical potential energy supplied by the battery. This energy is stored as long as the charge is held.

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8
Q

What is the equation for the energy stored in a capacitor

A

E = (1/2)QV = (1/2)CV^2 = (Q^2)/(2C)

The energy supplied to the capacitor is equal to the work done to move the charge against the pd across the plates.
W = QΔV
∴ E(supplied) = QV
The energy stored by a capacitor is equal to half the energy supplied to it (as 50% is lost to the load/internal resistance)
∴ E(stored) = (1/2)QV
Q = CV, ∴ E = (1/2)CV^2
V = Q/C, ∴ E = (Q^2)/(2C)

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9
Q

What is given by the area under a Q-V graph for a capacitor

A

The energy stored by the capacitor,
as E = (1/2)QV, due to the directly proportional relationship making this area a triangle. ((1/2)bh)

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10
Q

What is given by the gradient of a Q-V graph for a capacitor

A

Capacitance,
as C = Q/V

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11
Q

What is a dielectric

A

The insulator between the plates in a capacitor

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12
Q

What is permittivity (ε)

A

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

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13
Q

What causes the permittivity of a dielectric

A
  • The material contains lots of polar molecules, with +/- ends
  • When there is no charge, these molecules are arranged randomly
  • When there is a charge across the capacitors, the + end of the polar molecules is attracted to the negative plate (and vice versa)
  • This means the molecules align antiparallel with the electric field (Parallel but opposite direction)
  • Each of the molecules have their own electric field that now opposes the direction of the capacitors field
  • E-fields are vector fields, so these opposing fields reduce the overall electric field between the plates
    ∴ Larger ε = Larger opposing field, so it is therefore harder to generate a field within the material (increased Q needed to create field).
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14
Q

What is the effect of permittivity on the capacitance of a capacitor

A
  • A large Permittivity means a larger opposing field, reducing the overall electric field between the plates
  • This reduces the pd needed to transfer a unit charge across the capacitor, increasing the capacitance
    ∴ ε ∝ C
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15
Q

What is relative permittivity/dielectric constant (εr)

A

Ratio of the permittivity of a material to the permittivity of free space (dimensionless)

∴ εr = ε1/εo

εr: Relative permittivity
ε1: Permittivity of the material
εo: Permittivity of free space

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16
Q

What is the permittivity of free space (εo)

A

8.85x10^-12 (Fm^-1)

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17
Q

What is the equation for capacitance (relating to permittivity)

A

C = (Aεoεr)/d

εr: Relative permittivity
εo: Permittivity of free space
(∴ εoεr = ε1: Permittivity of the material)
A: Effective area of the plates
d: Distance between plates

18
Q

How would you investigate the effect of different factors on the capacitance of a capacitor

A

Set up 2 parallel plates , separated by a dielectric, connected to a capacitance meter (or determine capacitance from Q-V)
- Alter how much the plates overlap to change the effective area
- Use different materials to change εr
- Use multiple layers of the same material to alter the distance between the plates

(If C is known, Same set up can be used to determine ε1)

19
Q

How do capacitors charge

A
  • When connected to a dc power supply, electrons flow from the negative terminal of the power supply to the capacitor plate connected to it, giving it a negative charge
  • This negative charge repels electrons on the other plate, causing them to flow towards the positive terminal of the power supply
  • The same number of electrons are repelled from the + plate as are built up on the - plate, so equal but opposite charges build up (no charge can flow directly between the plates)

∴ Charge builds up across the capacitor

20
Q

What is the equation for the charge across a charging capacitor

A

Q = (Qo)(1- e^(-t/RC))
(Shows inverse of exponential decay)

Qo: Initial charge
t: time
R: Resistance
C: Capacitance
(RC = Time constant)

21
Q

What happens to the current in a circuit as a capacitor charges

A

The current is initially high, but as the charge builds up, electrostatic repulsion of electrons on the negative plate makes it harder for more electrons to be deposited. This means less electrons are repelled off of the positive plate, so the current falls.

∴ Current decreases as charge increases, until pd = V of battery

22
Q

What is the equation for the current across a charging capacitor

A

I = (Io)e^(-t/RC)
(Shows exponential decay)

Io: Initial current
t: time
R: Resistance
C: Capacitance
(RC = Time constant)

23
Q

What happens to the potential difference across a capacitor as it charges

A

The pd is initially 0, but as the charge increases, and C=Q/V for the capacitor (V ∝ Q), the potential difference increases. However, this only occurs until the pd is equal to V of the battery, as the pd can’t exceed that of the source.

∴ Pd increases as charge increases, until pd = V of battery

24
Q

What is the equation for the potential difference across a charging capacitor

A

V = (Vo)(1- e^(-t/RC))
(Shows inverse of exponential decay)

Vo: Initial pd
t: time
R: Resistance
C: Capacitance
(RC = Time constant)

25
Q

How do Q, I and V change as a capacitor charges

A

Q: Increases
I: Decreases
V: Increases

26
Q

How do capacitors discharge

A
  • Take out the battery and reconnect the circuit
  • Stored pd across the capacitor causes a current to flow (in the opposite direction to charging), as electrons on the negative terminal move to the positive terminal until there is no longer a charge stored

∴ Charge decreases until it reaches 0

27
Q

What is the equation for charge across a discharging capacitor

A

Q = (Qo)e^(-t/RC)
(Shows exponential decay)

Qo: Initial charge
t: time
R: Resistance
C: Capacitance
(RC = Time constant)

28
Q

What happens to the current in a circuit as a capacitor discharges

A

The current is initially high as electrons flow from the negative to the positive plate, but the current decreases as the charge across the capacitor decreases. This occurs as there are less electrons to move after some time.

∴ Current decreases until charge reaches 0

29
Q

What is the equation for charge across a discharging capacitor

A

I = (Io)e^(-t/RC)
(Shows exponential decay)

Io: Initial current
t: time
R: Resistance
C: Capacitance
(RC = Time constant)

30
Q

What happens to the potential difference across a capacitor as it discharges

A

The pd is initially high, but as the charge decreases, and C=Q/V for the capacitor (V ∝ Q), the potential difference decreases.

∴ pd decreases until charge reaches 0

31
Q

What is the equation for potential difference across a discharging capacitor

A

V = (Vo)e^(-t/RC)
(Shows exponential decay)

Vo: Initial pd
t: time
R: Resistance
C: Capacitance
(RC = Time constant)

32
Q

How do Q, I and V change as a capacitor charges

A

Q: Decreases
I: Decreases
V: Decreases

33
Q

What is the effect of charging/discharging a capacitor through a fixed resistor

A

The capacitor will take longer to charge/discharge, as 𝜏 = RC, so 𝜏 ∝ R

34
Q

How do you investigate a capacitor discharging through a fixed resistor

A
  • Charge a capacitor until it is fully charged
  • Remove the power source from the circuit and add a voltmeter/ammeter, with a data logger to record V and I
  • Close the switch and allow the capacitor to discharge
  • When the current = 0, use the computer to calculate the change in Q over time

This will show an exponential decrease in I, V and Q over time

35
Q

What are log-linear graphs

A

A plot where one axis is logarithmic

36
Q

What is the equation of the line, in the form y=mx+c, on a log-linear graph of ln(Q) against time
(same for any decreasing quantity - I, V(discharge), Q(discharge))

A

ln(Q) = (-1/RC)t + ln(Qo)

Q = (Qo)e^(-t/RC)
∴ ln(Q) = ln((Qo)e^(-t/RC))
∴ ln(Q) = ln(Qo) + (-t/RC)(ln(e))
∴ ln(Q) = (-1/RC)t + ln(Qo)

y: ln(Q)
m: -1/RC
x: t
c: ln(Qo)

37
Q

What is given by the area underneath a log-linear graph of ln(I)-t

A

Change in charge, ΔQ = IΔt

38
Q

What is the time constant of a capacitor (𝜏)

A

The time taken for the charge of a discharging capacitor to fall to ~37% of it’s original value, or the time for charging capacitor to reach ~63%

(Time at which t=RC)

39
Q

What is the equation for the time constant

A

𝜏 = RC

Time constant is the time when t = RC, as:
Q = (Qo)e^(-t/RC)
∴ If t = RC
Q = (Qo)e^-1
∴ Q/Qo = e^-1
∴ Q/Qo = 1/2.718
∴ Q/Qo = 0.37

40
Q

What it the ‘time to halve’ for a capacitor (T(1/2))

A

The time taken for Q, V or I of a discharging capacitor to decrease to half it’s initial value

41
Q

What is the equation for the ‘time to halve’ of a capacitor

A

T(1/2) = 0.69RC = 0.69𝜏

T(1/2) is the time at which Q=(1/2)Qo
Q = (Qo)e^(-t/RC)
∴ e^(-t/RC) = 1/2
∴ -t/RC = ln(1/2)
∴ -t/RC = ln(2^-1)
∴ t/RC = ln(2)
∴ t = ln(2)RC
ln(2) = 0.693…
∴ T(1/2) = 0.69RC
RC = 𝜏
∴ T(1/2) = 0.69𝜏

Physics (14 decks)

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