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PHYSICS, M6 2024 HSC 28

An electron gun fires a beam of electrons at 2.0 × 10\(^6\) m s\(^{-1}\) through a pair of parallel charged plates towards a screen that is 30 mm from the end of the plates as shown.

There is a uniform electric field between the plates of 1.5 × 10\(^4\) N C\(^{-1}\). The plates are 5.0 mm wide and 20 mm apart. The electron beam enters mid-way between the plates. \(X\) marks the spot on the screen where an undeflected beam would strike.

Ignore gravitational effects on the electron beam.
 

 

  1. Show that the acceleration of an electron between the parallel plates is 2.6 × 10\(^{15}\) m s\(^{-2}\).   (2 marks)

  2. Show that the vertical displacement of the electron beam at the end of the parallel plates is approximately 8.1 mm.   (2 marks)

  3. How far from point \(X\) will the electron beam strike the screen?   (3 marks)

Show Answers Only

a.   \(\text{Using}\ \ F=qE\ \ \text{and}\ \ F=ma:\)

  \(a=\dfrac{qE}{m} = \dfrac{1.5 \times 10^{4} \times 1.602 \times 10^{-19}}{9.109 \times 10^{-31}} = 2.6 \times 10^{15}\ \text{m s}^{-2} \)
 

b.  \(\text{Time of beam between the plates:}\)

\(\text{Horizontal velocity}\ (v)\ = 2 \times 10^{6}\ \text{m s}^{-1} \)

\(\text{Distance to screen}\ (s)\ = 5.0\ \text{mm}\ = \dfrac{5}{1000} = 0.005\ \text{m}\)

  \(t=\dfrac{s}{v} = \dfrac{0.005}{2 \times 10^{6}} = 2.5 \times 10^{-9}\ \text{s} \)
 

\(\text{Find vertical displacement at end of plates:}\)

  \(s=\dfrac{1}{2} at^{2} = 0.5 \times 2.6 \times 10^{15} \times (2.5 \times 10^{-9})^{2} = 0.008125\ \text{m}\ = 8.1\ \text{mm} \)
 

c.   \(\text{Find vertical velocity of beam when leaving the plates:}\)

  \(v=at=2.6 \times 10^{15} \times 2.5 \times 10^{-9} = 6.5 \times 10^{6}\ \text{m s}^{-1} \)
 

\(\text{Time for beam to hit screen:}\)

\(\text{Distance to screen}\ (s)\ = 30\ \text{mm}\ = \dfrac{30}{1000} = 0.03\ \text{m}\)

  \(t=\dfrac{s}{v} = \dfrac{0.03}{2 \times 10^{6}} = 1.5 \times 10^{-8}\ \text{s} \)
 

\(\text{Vertical displacement (from end of plates):}\)

  \(s=vt=6.5 \times 10^{6} \times 1.5 \times 10^{-8} = 0.0975\ \text{m} \)
 

\(\text{Distance from}\ X = 0.0081 + 0.0975 = 0.11\ \text{m} \)

Show Worked Solution

a.   \(\text{Using}\ \ F=qE\ \ \text{and}\ \ F=ma:\)

  \(a=\dfrac{qE}{m} = \dfrac{1.5 \times 10^{4} \times 1.602 \times 10^{-19}}{9.109 \times 10^{-31}} = 2.6 \times 10^{15}\ \text{m s}^{-2} \)
 

b.  \(\text{Time of beam between the plates:}\)

\(\text{Horizontal velocity}\ (v)\ = 2 \times 10^{6}\ \text{m s}^{-1} \)

\(\text{Distance to screen}\ (s)\ = 5.0\ \text{mm}\ = \dfrac{5}{1000} = 0.005\ \text{m}\)

  \(t=\dfrac{s}{v} = \dfrac{0.005}{2 \times 10^{6}} = 2.5 \times 10^{-9}\ \text{s} \)
 

\(\text{Find vertical displacement at end of plates:}\)

  \(s=\dfrac{1}{2} at^{2} = 0.5 \times 2.6 \times 10^{15} \times (2.5 \times 10^{-9})^{2} = 0.008125\ \text{m}\ = 8.1\ \text{mm} \)
 

c.   \(\text{Find vertical velocity of beam when leaving the plates:}\)

  \(v=at=2.6 \times 10^{15} \times 2.5 \times 10^{-9} = 6.5 \times 10^{6}\ \text{m s}^{-1} \)
 

\(\text{Time for beam to hit screen:}\)

\(\text{Distance to screen}\ (s)\ = 30\ \text{mm}\ = \dfrac{30}{1000} = 0.03\ \text{m}\)

  \(t=\dfrac{s}{v} = \dfrac{0.03}{2 \times 10^{6}} = 1.5 \times 10^{-8}\ \text{s} \)
 

\(\text{Vertical displacement (from end of plates):}\)

  \(s=vt=6.5 \times 10^{6} \times 1.5 \times 10^{-8} = 0.0975\ \text{m} \)
 

\(\text{Distance from}\ X = 0.0081 + 0.0975 = 0.11\ \text{m} \)

♦ Mean mark (c) 40%.

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)`.   (2 marks)
  1. Calculate the velocity of an electron as it reaches the anode.   (2 marks)
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 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.    Inaccuracies include:

  • 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.    Inaccuracies include:

  • 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)`  
  `=(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 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`

  `(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 (a) 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)

 
  `(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 (b) 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?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Downwards motion}\rule[-1ex]{0pt}{0pt}& \text{Upwards motion} \\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}>F_{\text{F}}\rule[-1ex]{0pt}{0pt}&F_{\text{E}}>F_{\text{F}}\\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}>F_{\text{F}}\rule[-1ex]{0pt}{0pt}& F_{\text{E}}>F_{\text{G}}+F_{\text{F}}\\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}=F_{\text{F}}\rule[-1ex]{0pt}{0pt}& F_{\text{G}}=F_{\text{E}} \\
\hline
\rule{0pt}{2.5ex}F_{\text{G}}=F_{\text{F}}\rule[-1ex]{0pt}{0pt}& F_{\text{E}}=F_{\text{G}}+F_{\text{F}} \\
\hline
\end{array}
\end{align*}

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. \(F_{\text{G}}=F_{\text{F}}\)).
  • 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. \(F_{\text{E}}=F_{\text{G}}+F_{\text{F}}\)).

\(\Rightarrow D\)


♦♦♦ Mean mark 26%.

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 and is therefore 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 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.
  2. 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.
  2. 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%.