SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

PHYSICS, M6 2020 VCE 3

Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).

Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
 

  1. Show that the velocity of an electron that travels straight through the aperture to point \(\text{Y}\) is given by  \( v_{0} \) = \( \dfrac{E}{B}\).   (1 mark)

  2. Calculate the magnitude of the velocity, \(v_0\), of an electron that travels straight through the aperture to point \(\text{Y}\) if  \(E\) = 500 kV m\(^{-1}\)  and  \(B\) = 0.25 T. Show your working.   (2 marks)

  3.  i. At which of the points – \(\text{X, Y}\), or \(\text{Z}\) – in Figure 2 could electrons travelling faster than \(v_0\) arrive?   (1 mark)

  4. ii. Explain your answer to part c.i.   (2 marks)

Show Answers Only

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  

 
b.    \(v_0=2 \times 10^6\ \text{ms}^{-1}\)

c.i.    Point \(\text{Z.}\)

c.ii.  When electrons travel faster than \(v_0\):

→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 

→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.

→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)

Show Worked Solution

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  
♦ Mean mark (a) 46%.

b.    \(v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}\)
 

c.i.    Point \(\text{Z.}\)
 

c.ii.  When electrons travel faster than \(v_0\):

→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 

→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.

→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)

♦ Mean mark (c.i.) 40%.
♦♦♦ Mean mark (c.ii.) 15%.
COMMENT: Students needed to include what would happen to the electric force.

PHYSICS, M6 2021 VCE 5

Figure 5 shows a stationary electron (e\(^{-}\)) in a uniform magnetic field between two parallel plates. The plates are separated by a distance of 6.0 × 10\(^{-3}\) m, and they are connected to a 200 V power supply and a switch. Initially, the plates are uncharged. Assume that gravitational effects on the electron are negligible.
 

  1. Explain why the magnetic field does not exert a force on the electron. Justify your answer with an appropriate formula.  (2 marks)

The switch is now closed.

  1. Determine the magnitude and the direction of any electric force now acting on the electron. Show your working.  (3 marks)

  1. Ravi and Mia discuss what they think will happen regarding the size and the direction of the magnetic force on the electron after the switch is closed.

    Ravi says that there will be a magnetic force of constant magnitude, but it will be continually changing direction.

    Mia says that there will be a constantly increasing magnetic force, but it will always be acting in the same direction.

    Evaluate these two statements, giving clear reasons for your answer.  (4 marks)

Show Answers Only

a.    → Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).

→ Electron is stationary (\(v=0\))

→ The force on the electron = \(0\).

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

→ Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).

c.    Both statements contain a correct assertion and incorrect assertion.

→ Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.

→ Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.

→ This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 

→ Mia was incorrect to say that the direction of the magnetic force on the electron would not change.

→ Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.

→ This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.

Show Worked Solution

a.    → Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).

→ Electron is stationary (\(v=0\))

→ The force on the electron = \(0\).
 

♦♦ Mean mark (a) 35%. 
COMMENT: Many students confused the forces of electric (not applicable here) and magnetic fields.

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

→ Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).
 

c.    Both statements contain a correct assertion and incorrect assertion.

→ Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.

→ Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.

→ This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 

→ Mia was incorrect to say that the direction of the magnetic force on the electron would not change.

→ Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.

→ This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.

♦♦♦ Mean mark (c) 18%.
COMMENT: Students need well planned responses to properly explain the physics principles required for 4 marks.

PHYSICS, M6 2023 HSC 12 MC

Figure \(\text{I}\) shows a positively charged particle accelerating freely from \(X\) to \(Y\), between oppositely charged plates. The change in the particle's kinetic energy is \(W\).

The distance between the plates is then doubled as shown in Figure \(\text{II}\) . The same charge accelerates from rest over the same distance from \(X\) to \(Y\).
 

What is the change in kinetic energy of the positively charged particle shown in Figure \(\text{II}\) ?

  1. \(W\)
  2. \(\dfrac{W}{2}\)
  3. \(\sqrt{W}\)
  4. \(2 W\)

Show Answers Only

\(B\)

Show Worked Solution

→ In Figure \(\text{I}\):  \( W= qEd\)  and  \( E=\dfrac{1000V}{d} \)

\(W=\dfrac{1000qVd}{d}=1000qV\)

→ In Figure \(\text{II}\):  \( \Delta \text{KE} =qEd\)  and  \( E=\dfrac{1000V}{2d} \)

\(\Delta \text{KE}=\dfrac{1000qVd}{2d}= \dfrac{1}{2} \times 1000qV= \dfrac{W}{2}\)

\(\Rightarrow B\)

♦ Mean mark 55%.

PHYSICS, M6 EQ-Bank 22

An 'electron gun' like that used by JJ Thomson is shown.
 

Electrons leave the cathode and are accelerated towards the anode.

  1. Show that the acceleration of the electrons as they just leave the cathode is  `4 × 10^(16) \ text{m s}^(-2)`.
  1. Calculate the velocity of an electron as it reaches the anode.
Show Answers Only
  1. `4 xx10^(16)  text{m s}^(-2)`
  2. `4 xx10^7  text{m s}^(-1)`
Show Worked Solution

a.    → The force on the electrons is given by  `F=qE`.

→ The electric field strength is given by  `E=(V)/(d)`:

`F` `=(Vq)/(d)`  
`a` `=(F)/(m)`  
  `=(Vq)/(dm)`  

 
→ Calculate the acceleration:

`a` `=(Vq)/(dm)`  
  `=(5000 xx1.602 xx10^(-19))/(0.02 xx9.109 xx10^(-31))`  
  `=4 xx10^(16)  text{m s}^(-2)`  

 

b.  → Using kinematic equations:

`v^(2)` `=u^(2)+2as`  
`v` `=sqrt(2as),\ \ \ (u=0)`  
  `=sqrt(2xx4xx10^(16)xx0.02)`  
  `=4 xx10^7  text{m s}^(-1)`   

PHYSICS, M6 EQ-Bank 26

The diagram shows a stationary electron in a magnetic field. The magnetic field is surrounded by two parallel plates separated by a distance of `5.0 × 10^(-3) \ text {m}` and connected to a power supply and a switch.
 

The switch is initially open. At a later time the switch is closed.

Analyse the effects of the magnetic and electric fields on the acceleration of the electron both before and immediately after the switch is closed. In your answer, include calculation of the acceleration of the electron immediately after the switch is closed.

Show Answers Only

“

Show Worked Solution

→ Before the switch is closed, there is no electric field and so no electric force causing the electron to accelerate. As the electron is stationary, the magnetic field has no effect on it.

→ Once the switch is closed, the electric field causes the electron to accelerate down the page (towards the positive plate).

→ Calculating the acceleration of the electron immediately after the switch is closed:

`E` `=(V)/(d)`  
  `=(100)/(0.005)`  
  `=20\ 000\ text{V m}^(-1)`  
`F` `=Eq`  
  `=20\ 000 xx1.602 xx10^(-19)`  
  `=3.2 xx10^(-15)  text{N}`  
`a` `=(F)/(m)`  
  `=(3.2 xx10^(-15))/(9.109 xx10^(-31))`  
  `=3.5 xx10^(15)  text{m s}^(-2)  text{down the page.}`  

 
→ Now that the electron is moving, the magnetic field exerts a force on it towards the right (using the right hand palm rule).

→ The force and acceleration on the electron will increase due to its increasing velocity.

PHYSICS, M6 EQ-Bank 8 MC

A positively-charged ion travelling at 250 ms ¯1 is fired between two parallel charged plates, `M` and `N`. There is also a magnetic field present in the region between the two plates. The direction of the magnetic field is into the page as shown. The ion is travelling perpendicular to both the electric and the magnetic fields.
 

The electric field between the plates has a magnitude of 200 V m ¯1. The magnetic field is adjusted so that the ion passes through undeflected.

What is the magnitude of the adjusted magnetic field, and the polarity of the `M` terminal relative to the `N` terminal?
 

Show Answers Only

`A`

Show Worked Solution

→ The ion passes through undeflected, so the magnitude of the force it experiences due to the magnetic field is equal to the magnitude of the force it experiences due to the electric field. 

→ As the ion travels perpendicular to the magnetic field:

`F` `=qvB`  
`Eq` `=qvB`  
`B` `=(E)/(v)`  
  `=(V)/(d)((1)/(v))`    `(E=(V)/(d))`
  `=(200)/(250)`  
  `=0.8  text{T}`  

 
→ Using the right hand palm rule, the magnetic field exerts a force up the page on the positive ion. The electric field must therefore exert a force down the page on the ion.

→ `M` must be positive relative to `N`.

`=>A`

PHYSICS, M6 2015 HSC 24

A part of a cathode ray oscilloscope was represented on a website as shown.
 

Electrons leave the cathode and are accelerated towards the anode.

  1. Explain why the representation of the path of the electron between the deflection plates is inaccurate.   (3 marks)
  1. Calculate the force on an electron due to the electric field between the cathode and the anode.   (2 marks)
  1. Calculate the velocity of an electron as it reaches the anode.   (2 marks)
Show Answers Only

a.   There are a number of inaccuracies:

→ The electron is shown to travel towards the negative plate. This is inaccurate as the electron is a negatively charged particle so will be attracted towards the positive plate.

→ The electron is shown to only deflect once it has travelled halfway through the plates. This is inaccurate as the plate will deflect the electron immediately after it enters the uniform electric field. 

→ The electron is shown to follow a straight path. This is inaccurate as the electron has a constant acceleration towards the positive plate and a constant horizontal velocity, it will follow a parabolic path.

b.   `F=4xx10^(-14)  text{N}`

c.   `v=4.19 xx10^(7)  text{ms}^(-1)`

Show Worked Solution

a.   There are a number of inaccuracies:

→ The electron is shown to travel towards the negative plate. This is inaccurate as the electron is a negatively charged particle so will be attracted towards the positive plate.

→ The electron is shown to only deflect once it has travelled halfway through the plates. This is inaccurate as the plate will deflect the electron immediately after it enters the uniform electric field. 

→ The electron is shown to follow a straight path. This is inaccurate as the electron has a constant acceleration towards the positive plate and a constant horizontal velocity, it will follow a parabolic path.
 

b.   Using `E=(F)/(q)`  and  `E=(V)/(d):`

`(F)/(q)` `=(V)/(d)`  
`F` `=(Vq)/(d)`  
`F` `=(5000 xx1.6 xx10^(-19))/(0.02)`  
  `=4xx10^(-14)  text{N}`  
  

c.    `a` `=(F)/(m)`
    `=(4xx10^(-14))/(9.1 xx10^(-31))`
    `=4.4 xx10^(16)  text{ms}^(-1)`

 

`v^(2)` `=u^(2)+2as`  
`:.v` `=sqrt(2xx4.4 xx10^(16)xx0.02)`  
  `=4.19 xx10^(7)  text{ms}^(-1)`  

♦♦♦ Mean mark (c) 20%.

PHYSICS, M6 2016 HSC 3 MC

A region of space contains a constant magnetic field and a constant electric field.

How will these fields affect an electron that is stationary in this region?

  1. Both fields will exert a force.
  2. Neither field will exert a force.
  3. Only the electric field will exert a force.
  4. Only the magnetic field will exert a force.
Show Answers Only

`C`

Show Worked Solution

→ Magnetic fields only exert a force on moving charged particles while electric fields exert a force on all charged particles moving or stationary.

`=>C`


♦♦ Mean mark 35%.

PHYSICS, M6 2017 HSC 30

In a thought experiment, a proton is travelling at a constant velocity in a vacuum with no field present. An electric field and a magnetic field are then turned on at the same time.

The fields are uniform in magnitude and direction and can be considered to extend infinitely. The velocity of the proton at the instant the fields were turned on is perpendicular to the fields.
 

Analyse the motion of the proton after the fields have been turned on.   (4 marks)

Show Answers Only

→ Using the right hand palm rule, the magnetic field exerts a force out of the page, perpendicular to the velocity of the proton.

→ So, the magnetic field causes the proton to undergo circular motion, initially moving out of the page and continuing in an anti-clockwise direction as viewed from the right.

→ The electric field exerts a constant force to the left on the proton.

→ This causes the proton to accelerate towards the left. The resultant motion will be a helix (vector sum of motion) that extends to the left, with the distance between adjacent spirals increasing.

→ The helix will decrease in radius as the proton loses kinetic energy and hence speed as it radiates electromagnetic waves. This occurs because the proton is an accelerating charge.

Show Worked Solution

→ Using the right hand palm rule, the magnetic field exerts a force out of the page, perpendicular to the velocity of the proton.

→ So, the magnetic field causes the proton to undergo circular motion, initially moving out of the page and continuing in an anti-clockwise direction as viewed from the right.

→ The electric field exerts a constant force to the left on the proton.

→ This causes the proton to accelerate towards the left. The resultant motion will be a helix (vector sum of motion) that extends to the left, with the distance between adjacent spirals increasing.

→ The helix will decrease in radius as the proton loses kinetic energy and hence speed as it radiates electromagnetic waves. This occurs because the proton is an accelerating charge.


♦ Mean mark 51%.

PHYSICS, M6 2018 HSC 12 MC

The diagram shows electrons travelling in a vacuum at  `2 × 10^6 \ text{m s}^(-1)`  between two charged metal plates  `1 × 10^-3\ text{m}`  apart.
 

A magnetic field is to be applied to make the electrons continue to travel in a straight line.

What is the magnitude and direction of the magnetic field that is to be applied?

  1. `5 × 10^-1 \ text {T}`  into the page
  2. `5  × 10^-1 \ text {T}`  out of the page
  3. `1 × 10^6 \ text {T}`  into the page
  4. `1 × 10^6 \ text {T}`  out of the page
Show Answers Only

`A`

Show Worked Solution

→ There will be an upwards force (towards the positive plate) on the electrons due to the electric field.

→ Using the right hand palm rule, the magnetic field must be into the page to produce a downwards force and balance the upwards force.

→ Since the magnitudes of electric and magnetic force are equal:

`qE` `=qvB`  
`B` `=(E)/(v)`  
  `=((1000)/(1xx10^(-3)))/(2xx10^(6))`  
  `=5xx10^(-1)\ text{T}`  

 
`=>A`


♦ Mean mark 48%.

PHYSICS, M6 2020 HSC 34

A charged particle, `q_1`, is fired midway between oppositely charged plates `X` and `Y`, as shown in Figure 1. The voltage between the plates is `V` volts.

The particle strikes plate `Y` at point `P`, a horizontal distance `s` from the edge of the plate. Ignore the effect of gravity.
 

Plate `Y` is then moved to the position shown in Figure 2, with the voltage between the plates remaining the same.

An identical particle, `q_2`, is fired into the electric field at the same velocity, entering the field at the same distance from plate `X` as `q_1`.

  1. Compare the work done on `q_1` and `q_2`.   (3 marks)
  1. Compare the horizontal distances travelled by `q_1` and `q_2` in the electric field.   (3 marks)
Show Answers Only

a.   The work done on `q_2` is `(3)/(2)` times that done on `q_1`.

b.   The horizontal distance travelled by `q_2` is `sqrt(6)` times that of `q_1`.

Show Worked Solution

a.   Work done on `q_1:`

`W=qE((d)/(2))` … (1)

Substitute  `E=(V)/(d)`  into (1)

`W=(qV)/2`

 
Work done on `q_2:`

`W=qE((3d)/(2))` … (1)

Substitute  `E=(V)/(2d)`  into (1)

`W=(3qV)/4`

Divide  (2) `-: ` (1):

`(W_(2))/(W_(1))=(((3qV)/(4)))/(((qV)/(2)))=(3)/(2)`

→ The work done on `q_2` is `3/2` times that done on `q_1`.
 

♦ Mean mark part 47%.

b.   Find the vertical acceleration of `q_1:`

`F` `=qE=(qV)/(d)`
`a` `=(F)/(m)=(qV)/(dm)` … (1)

 

Similarly for `q_2:` 

`F` `=qE=(qV)/(2d)`
`a` `=(F)/(m)=(qV)/(2dm)` … (2)

 
→ the initial horizontal velocity is the same for both `q_1` and `q_2`

→ the horizontal distance travelled by both is proportional to the time taken for them to reach the plate.

→ `s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)=(1)/(2)a_(y)t^(2)`  (initial vertical velocity = 0 for both)
 

Time for `q_1` to reach plate `Y:` 

`s_(y)` `=(1)/(2)a_(y)t^(2)`  
`(d)/(2)` `=(1)/(2)((qV)/(dm))t^(2)`    from (1)
`t^(2)` `=(d^(2)m)/(qV)`  
`t` `=sqrt((d^(2)m)/(qV))` …   (3)

 
Time taken for `q_2` to reach plate `Y:` 

`s_(y)` `=(1)/(2)a_(y)t^(2)`  
`(3d)/(2)` `=(1)/(2)((qV)/(2dm))t^(2)`    from (2)
`t^(2)` `=(6d^(2)m)/(qV)`  
`t` `=sqrt((6d^(2)m)/(qV))` …   (4)

 
Divide  (4) ÷ (3):

`(t_(2))/(t_(1))=(sqrt((6d^(2)m)/(qV)))/(sqrt((d^(2)m)/(qV)))=sqrt(6)`

→ The time taken for `q_2` to reach plate `Y` is `sqrt(6)` times that taken for `q_1`.

→ The horizontal distance travelled by `q_2` is `sqrt(6)` times that of `q_1`.

♦ Mean mark part 43%.

PHYSICS M6 2022 HSC 18 MC

A charged oil droplet was observed between metal plates, as shown.
 


 

While the switch was open, the oil droplet moved downwards at a constant speed. After the switch was closed, the oil droplet moved upwards at the same constant speed.

Assume that the only three forces that may act on the oil droplet are the force of gravity, the force due to the electric field and the frictional force between the air and the oil droplet. The magnitudes of these forces are `F_G` (due to gravity), `F_E` (due to the electric field) and `F_F` (due to the frictional force).

Which row of the table shows all the forces affecting the motion of the oil droplet in the direction indicated, and the relationship between these forces?
 

Show Answers Only

`D`

Show Worked Solution

→ For the droplet to move at a constant speed, the net force acting on it must be zero.

→ For the downwards motion, this means the downwards gravitational force is equal in magnitude to the upwards frictional force (i.e. `F_(G)=F_(F)`).

→ For the upwards motion, this means the upwards electric force is equal in magnitude to the sum of the downwards frictional and gravitational forces (i.e. `F_(E)=F_(G)+F_(F)`).

`=>D`


♦♦♦ Mean mark 26%.

PHYSICS M6 2022 HSC 12 MC

The diagram shows a region in which there are uniform electric and magnetic fields. A positively charged particle moves in the region at constant velocity.
 


 

What is the direction of the particle's velocity?

  1. Up the page
  2. Down the page
  3. To the left
  4. To the right
Show Answers Only

`D`

Show Worked Solution

→ For the particle to have a constant velocity, the net force acting on it must be zero.

→ The electric field exerts a downwards force on the particle.

→ The magnetic field must exert an upwards force on the particle.

→ Using the right hand palm rule, the direction of the particle’s velocity is to the right.

`=>D`


♦ Mean mark 52%.

PHYSICS M6 2022 HSC 3 MC

A radioisotope emits radiation which is deflected by an electric field, as shown.
 


 

What type of radiation is this?

  1. Alpha
  2. Gamma
  3. Beta positive (positron)
  4. Beta negative (electron)
Show Answers Only

`D`

Show Worked Solution

The radiation experiences a force of attraction towards the positive plate.

→ it must be negatively charged.

`=>D`

PHYSICS, M6 2019 HSC 16 MC

The diagram shows the trajectory of a particle with charge `q` and mass `m` when fired horizontally into a vacuum chamber, where it falls under the influence of gravity.
 

The horizontal distance, `d`, travelled by the particle is recorded.

The experiment is repeated with a uniform vertical electric field applied such that the particle travels the same horizontal distance, `d`, but strikes the upper surface of the chamber.
 

What is the magnitude of the electric field?

  1. `mgq`
  2. `2mgq`
  3. `(mg)/q`
  4. `(2mg)/q`
Show Answers Only

`D`

Show Worked Solution

→ As the particle travels the same distance in both scenarios the magnitude of the force it experiences is the same.

→ Initially, the only force acting on the particle is its weight, `F=mg.`

→ When the electric field is on, the particle experiences an upwards force due to the electric field, and a downwards force due to gravity,  `F=qE-mg.`

→ As these forces have the same magnitude:

`qE-mg` `=mg`  
`qE` `=2mg`  
`E` `=(2mg)/(q)`  

 
`=>D`


♦ Mean mark 42%.

PHYSICS, M6 2020 HSC 10 MC

An electron travelling in a straight line with an initial velocity, `u`, enters a region between two charged plates in which there is an electric field causing it to travel along the path as shown.
 

A magnetic field is then applied causing a second electron with the same initial velocity to pass through undeflected.

Which row of the table shows the directions of the electric and magnetic fields when the second electron enters the region between the plates?
 

Show Answers Only

`B`

Show Worked Solution

The bottom plate is positively charged as it attracts the electron.

→ Electric field is towards the top of the page.

Using the right hand palm rule, the magnetic field must be out of the page to produce an upwards force on the electron.

`=>B`


♦ Mean mark 50%.

PHYSICS, M6 2021 HSC 30

In an experiment, a proton accelerates from rest between parallel charged plates. The spacing of the plates is 12 cm and the proton is initially positioned at an equal distance from both plates, as shown. Ignore the effect of gravity.
 

  1. The electrical potential energy of the proton is recorded in the following graph for the first 5 cm of its motion.

   On the graph, sketch the corresponding kinetic energy of the proton over the same distance.   (2 marks)
 
 

  1. The experiment is repeated using an electron in the place of the proton.

Explain how the motion of the electron would differ from that of the proton.    (3 marks)

Show Answers Only

a.   
           
        
b.   Motions differs from proton motion:

→ The electron will move in the opposite direction to the proton as the sign of its charge is opposite.

→ The electron will experience a force equal in magnitude to the proton as its charge has the same magnitude.

→ Since the mass of the electron is significantly smaller, it will experience a much greater acceleration compared to the proton.

Show Worked Solution

a.   
           
        
  

b.   Motions differs from proton motion:

→ The electron will move in the opposite direction to the proton as the sign of its charge is opposite.

→ The electron will experience a force equal in magnitude to the proton as its charge has the same magnitude.

→ Since the mass of the electron is significantly smaller, it will experience a much greater acceleration compared to the proton.

PHYSICS, M6 2021 HSC 18 MC

An evacuated chamber contains a pair of parallel plates connected to a power supply and a switch which is initially closed.

A positively charged mass (•) falls within the chamber, under the influence of gravity, from the position shown.
 

When the mass has fallen half the height of the chamber, the switch is opened.

Which of the following correctly shows the trajectory of the mass?
 

 

Show Answers Only

`A`

Show Worked Solution

The mass experiences a constant force to the right due to the electric field and a constant force downwards due to gravity.

This results in a net force pointing diagonally down and to the right. The mass travels in the direction of this net force.

When the switch is opened the mass falls only under the influence of gravity, following a parabolic path.

`=>A`


♦ Mean mark 35%.

PHYSICS, M6 2021 HSC 2 MC

A positively charged particle is moving at velocity, `v`, in an electric field as shown.
 

What is the direction of the force acting on the particle due to the electric field?

  1. Into the page
  2. Out of the page
  3. Up the page
  4. Down the page
Show Answers Only

`C`

Show Worked Solution

Force of positive charge → direction of electric field

`=>C`


♦ Mean mark 38%.