March 2013

8.1 Stores (electric) charge (energy).

OR

Releases (stored) charge instantly/very fast.

OR

Releases (stored) charge instantly/very fast.

A high-resistance light bulb and an uncharged parallel plate capacitor are connected in series with a 12V battery and a switch

8.2 The brightness of the bulb decreases (gradually) until it stops glowing (dies).

OR

The bulb glows dimmer until it stops glowing (dies).

This is bacause as the capacitor gets charged, it develops an increasing opposing potential difference to the very battery that is charging it. This opposition gradually decreases the current in the cicuit, until the currecnt becomes zero.

OR

The bulb glows dimmer until it stops glowing (dies).

This is bacause as the capacitor gets charged, it develops an increasing opposing potential difference to the very battery that is charging it. This opposition gradually decreases the current in the cicuit, until the currecnt becomes zero.

The capacitor is NOW fully charged.

8.3.1 0 V (zero volts)

no current flowing, therefore no potential difference.

no current flowing, therefore no potential difference.

8.3.2 12V

8.4.1

• The direction is from positive plate to negative plate.

• The electric field is UNIFORM, which means of constant strength and direction.

• The direction is from positive plate to negative plate.

• The electric field is UNIFORM, which means of constant strength and direction.

8.4.2

Since the electric field is uniform, the electron would experience the same size (constant) force ANYWHERE in between the plates, even close to the negative plate.

As a result of the constant force, the acceleration of the electron would be constant as well. (Newton's Second Law)

Since the electric field is uniform, the electron would experience the same size (constant) force ANYWHERE in between the plates, even close to the negative plate.

As a result of the constant force, the acceleration of the electron would be constant as well. (Newton's Second Law)

Calculate the:

8.5.1 5,4mm - 3,8mm = 1,6mm

8.5.2

**(a) Calculate the work involved when an electron moves from rest from the negative plate to the positive plate.**

W = VQ

W = 12(1,6 x 10^{-19})

W = 1,92 x 10^{-18}J

OR

W = FΔcosΘ

W = (3,56 x 10^{-16})(5,4 x 10^{-3})(1)

W = 1,92 x 10^{-18}J

**NOTE:** If the electron is to be moved TO the negative plate, work must be done to it by an external source. (a kind of "repulsion" to be overcome.)

If the electron is to be moved to the positive plate, an external source is not required, since this is a natural movement.

**(b) Calculate the velocity of the electron when it reaches the positive plate. (Work energy theorem)**

**(c) How would this magnitude of work calculated in (a), be affected if the distance between the plates was halved?**

The work done would be the same.

None of the variables in W = VQ would have changed. Only d changed, and d is not part of this equation.

**NOTE:**

But E would have increased

E =^{V}/_{d}

Hence F would have increased since F = QE (same Q, greater E)

Hence acceleration would have increased since larger F.

**(d) How would the final velocity of the electron be affected if the distance between the plates was halved.**

No change.

No change to variables Ek, m and v_{i} in the equation. Hence v_{f} would be the same.

W = VQ

W = 12(1,6 x 10

W = 1,92 x 10

OR

W = FΔcosΘ

W = (3,56 x 10

W = 1,92 x 10

If the electron is to be moved to the positive plate, an external source is not required, since this is a natural movement.

The work done would be the same.

None of the variables in W = VQ would have changed. Only d changed, and d is not part of this equation.

But E would have increased

E =

Hence F would have increased since F = QE (same Q, greater E)

Hence acceleration would have increased since larger F.

No change.

No change to variables Ek, m and v

Nov 2012

a 1 000Ω resistor. The emf of the battery is 12 V. Ignore the internal resistance of the battery and the ammeter.

8.1

8.2 12 V

The capacitor has a capacitance of 120µF and the space between its plates is filled with air.

8.3

After discharging the capacitor, it is connected in the same circuit to a resistor of HIGHER resistance and switch

(Write down INCREASES, DECREASES or REMAINS THE SAME.)

8.4.1 Decreases

8.4.2 Increases

8.5.1

8.5.2

March 2010

10.1 Dielectric

Distance between plates

Distance between plates

10.2.1

10.2.2

Calculate the area A of the plates as length x breadth by convertng the centimeters to meters.

Calculate the area A of the plates as length x breadth by convertng the centimeters to meters.

Nov 2010

9.1 The ratio of the (amount of) charge (transferred) to the (resulting) potential difference.

9.2

Calculate the area A by converting the mm into m.

Calculate the area A by converting the mm into m.

The circuit diagram below shows the ABOVE CAPACITOR, initially uncharged, connected in series to a resistor, an ammeter of negligible resistance and a source with an emf of 12 V. The internal resistance of the battery is negligible.

Switch

9.3

The capacitor is now fully charged.

9.4

9.5

Then

Then

March 2009

10.1

Convert the 40cm^{2} into meter squared. (4 x 10^{-3} m^{2})

Convert the 40cm

Explain your answer in terms of physics principles and the charge stored in the capacitor. (NO calculations needed.) (3)

10.2 half

Half the area will store half the amount of charge

OR

C ∝ A

and C ∝ Q,

thus C is halved

Half the area will store half the amount of charge

OR

C ∝ A

and C ∝ Q,

thus C is halved

10.3 net charge = 0 C

10.4 Discharges almost instantly to deliver flash light

Nov 2009 Unused

10.1 Discharges very fast when touched and can cause an electric shock that can be fatal

10.2.1

10.2.2

The second step is Q = CV

The second step is Q = CV

10.3.1 Increases

10.3.2 Decreases

Nov 2008

10.1 Electric force experienced per positive charge placed at the point

OR

A point /space where a charge will experience an electric force

OR

A point /space where a charge will experience an electric force

10.2 Negative

• Negative ink droplets deflect away from B

• Are attracted towards A

• B repels P

• like charges repel

• Negative ink droplets deflect away from B

• Are attracted towards A

• B repels P

• like charges repel

10.3

The plates A and B are 6,4 x 10

magnitude 1,5 x 10

10.4

Prep Paper 2008

11.1

This capacitor is connected across a 250 V source as shown below.

11.2

11.3 Increase the potential difference

The term "design" means the area A of the plates, and the distance d between the plates. Therefore you must not alter A or d.

The term "design" means the area A of the plates, and the distance d between the plates. Therefore you must not alter A or d.

11.4 dielectric

11.5 The capacitor stores charge. This large amount of charge can cause shock to the body.

Exemplar 2008

After closing the switch, the learner takes the ammeter readings every 20 seconds. The table below shows the results obtained during the investigation.

[HINT: The graph is not a straight-line.] (5)

10.1 **Graph of electric current versus time**

after 30 s. (1)

10.2 56µA

10.3 As the **potential difference across the plates of the capacitor
increases** during charging, the **potential difference of the battery is
opposed**, causing the current in the circuit to gradually decrease.

A capacitor is rated 9 V, 50 µF.

10.4

10.5 Any one

• Supply electrical energy faster

• Can be recharged almost indefinitely

• No spilling of dangerous chemicals

• Supply electrical energy faster

• Can be recharged almost indefinitely

• No spilling of dangerous chemicals

10.6 The high voltage across plates can cause electric shock or even death
when the capacitor discharges.

Additional Exemplar 2008

10.1 As the capacitor charges, the direct current decreases and eventually becomes zero when the capacitor is fully charged.

• Two connecting wires

• A whole sheet of aluminium foil of area 0,2m

Use the following steps as guidance in your design:

10.2.1

10.2.2

10.2.3 Double the distance between the plates / Increase the distance between the plates to 8,86 mm

10.3.1 Batteries store energy in chemical reactions

Capacitors store energy in electric fields

Capacitors store energy in electric fields

10.3.2 Any one:

Chemicals e.g. acid or heavy metals can leach into soil and groundwater.

Plastic casings can pollute environment.

Chemicals e.g. acid or heavy metals can leach into soil and groundwater.

Plastic casings can pollute environment.

'Capacitors cannot function without batteries - they need a source of energy. On the other hand, batteries don't need capacitors.'

Briefly explain why this is a valid statement.

10.3.3 Capacitors need a source of energy e.g. batteries to obtain charge.

Batteries produce their own energy (electricity) from chemical reactions inside the battery.

Batteries produce their own energy (electricity) from chemical reactions inside the battery.