Chapter 23 Capacitors

Question 1

What is a capacitor?

Answer 1

A device designed to store charge. It consists of two conductors insulated from each other

Two metal plates placed near each other form a capacitor

Question 2

What happens when a capacitor is connected to a battery?

Answer 2

One of the two conductors gains an electron form the battery while the other conductor loses electrons to the battery. A equal number of electrons is transferred so each plate becomes equal and opposite in charge. +Q and -Q

Question 3

What is the capacitance of a capacitor? The unit of capacitance?

Answer 3

The capacitance C of a capacitor is defined as the amount of charge stored per unit of potential difference.

C = Q/V

The unit of capacitance is the farad (F) = 1 x 10^-6

C can also be called constant of proportionality.

Question 4

Why is energy stored when a capacitor is charged?

Answer 4

This is because electrons are forced onto one of its plates and taken off the other plate. This energy is stored in the capacitors as electric potential energy

Question 5

How can you charge a capacitor at constant current.

Answer 5

When the switch is close the charge Q can be calculated using Q = IT at any given time after the switch closes. where I is the current

Question 6

What is the relationship between charge and PD? What will it show when plotted on a graph.

Answer 6

Q is proportional to pd V, the charge stored per volt is constant so on a graph it should be a straight line.

Question 7

What is a farad?

Answer 7

One farad is the amount of capacitance when one coloumb is charged for one volt

Question 8

Explain the charging of a capacitor in terms of electron flow

Answer 8

Electrons will flow from the negative terminal of the cell to the negative plate of the capacitor.

The excess electrons on the negative plate are going to repel electrons off the positive plate of the capacitor, which will move towards the positive terminal of the cell.

Overall the plates will acquire an equal and opposite charge.

Question 9

Describe the discharging of a capcitor and what function it provides

Answer 9

When a capacitor is connected to a circuit, the excess electrons flow from the negative plates to the positive plates. The charge of the capacitor will decrease. This charge on the capacitor decreases.

Question 10

How do you add capacitors in series?

Answer 10

1/CT = 1/C1 + 1/C2

In a calculator use brackets raise to power -1

1/CT = (1/C1 + 1/C2)^-1

To note they are exactly the opposite to adding resistors

Question 11

How do you add capacitors in parallel

Answer 11

CT = C1 + C2

Question 12

Provide a brief description of key facts of a parallel circuit.

Answer 12

• The voltage is the same across each parallel branch

-The charge is shared as current is shared

QT = Q1 + Q2
CT = C1 + C2
CTV = C1V + C2V - Potential difference is cancelled out

Question 13

Provide a brief description of key facts of a series circuit.

Answer 13

• Each capacitor will experience the same current so it will aquire the same charge
• The p.d is shared by kirchoff’s second law

QT = Q1 = Q2
VT= V1 + V2

Question 14

What is the equation for the voltage across a capacitor at time t after it begins to discharge?

Answer 14

Vοe^(-t/CR)

Where c is capcitor/tance and R is resistor/tance

Question 15

What is the equation for the Charge across a capacitor at time t after it begins to discharge?

Answer 15

Qοe^(-t/CR)

Where c is capcitor/tance and R is resistor/tance

Question 16

What is the equation for the Current across a capacitor at time t after it begins to discharge?

Answer 16

Iοe^(-t/CR)

Where c is capcitor/tance and R is resistor/tance

Question 17

A 500 microFarad capacitor is charged up to 6.0
V. It then discharges through a 500 kiloohm
resistor. Find the p.d. across the capacitor 150
seconds after it begins to discharge!

Answer 17

Vοe^(-t/CR)
Where c is capcitor/tance and R is resistor/tance

6 x e^(-150/(500x10^-6 x 500x10^-6))

Question 18

What is the constant property ratio of exponentials

Answer 18

V1/V0 = V2/V1 = V3/V2

Question 19

What is the equation for the voltage across a capacitor at time t after it begins to discharge? rearranged for t

Answer 19

Vοe^(-t/CR)

V/Vο = e^(-t/CR)

ln V/Vο = ln e^(-t/CR)

ln V/Vο = -t/CR

• ln V/Vο = t/CR

ln Vo/V = t/CR

t = CR x ln Vo/V

-

Question 20

The charging and discharging of aa capcitor depends on? What is time constant equation?

Answer 20

the resistance and capacitance

T = CR or T = Q/I ot Q = It

in seconds

Question 21

What happens too the equation when time constant is introduced?

Answer 21

Vοe^(CR/CR)

Vοe^-1

At time constant, the pd will drop to 37% of its original value, so will the charge and so will the current

Question 22

How can tie constant be figured out through a graph?

Answer 22

By calculating 37% of the voltage initial value. then using T = CR

Question 23

Describe voltage across the circuit

Answer 23

Initially the voltage is zero

As soon as a circuit is connected the capacitor starts charging.

The voltage across the capacitor will increase while the voltage across the resistor will start decreasing exponentially.

Vο = VR + VC

Vοe^(-t/CR) + VC

rearrange for voltage across capacitor VC = Vo - Vοe^(-t/CR)
VC = Vo (1 - e^(-t/CR) )

Charge across the capacitor: QC = Qo (1 - e^(-t/CR) )

while Voltage across resitor = Vοe^(-t/CR)

Current decreases as the capacitor is charging I = Iοe^(-t/CR)

Question 24

How can energy stored in a capcitor be calculated?

Answer 24

E = 1/2QV or 1/2CV^2

Question 25

What is used to measure the energy stored in a charged capacitor?

Answer 25

A joule meter

E =

Question 26

What is the potential difference between a thundercloud and the ground?

Answer 26

V=Ed where E is the electric field strength and d is the height of the thundercloud above the ground.

For a thundercloud carrying a constant charge Q, the energy
stored = 1/2QV = 1/2QEd.
If the thundercloud is forced by winds to rise up to a new height d’,
then the energy stored = 1/2QEd’.
Because the electric field strength is unchanged (because it depends
on the charge per unit area; see Topic 22.2), then the increase in
the energy stored =1/2QEd’ - 1/2QEd = 1/2QEΔd, where Δd = d’ - d.

The energy stored increases because work is done by the force of
the wind. It has to overcome the electrical attraction between the
thundercloud and the ground to make the charged thundercloud
move away from the ground.
The insulating property of air breaks down if it is subjected to an
electric field stronger than about 300kVm-1. Prove for yourself that,
for every metre rise of the thundercloud carrying a charge of 20C,
the energy stored would increase by 3 MJ. At a height of 500m, the
energy stored would be 1500 MJ.

Question 27

Difference Between Energy Supplied by the Battery and Energy Stored in the Capacitor

Answer 27

The energy supplied by the battery is the energy it delivers to charge up the capacitor. This charge is the same as the energy stored in the capacitor

Once the capacitor is fully charged, the battery stops delivering energy, and the capacitor holds onto the stored energy in the form of electric potential energy.

Question 28

Final Charge Stored by Each Capacitor:

Answer 28

When the switch is connected to Y, the charge is shared between the two capacitors in proportion to their capacitance. We can use the principle of charge conservation to find the final charge on each capacitor.

The total charge remains constant, so:

Q initial = Q final
Q1
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