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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 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)
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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%.