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PHYSICS, M4 EQ-Bank 5 MC

A proton experiences a force of \(3.2 \times 10^{-15}\ \text{N}\) when placed between two charged parallel plates that are separated by 2 mm.

What is the potential difference between the plates?

  1. \(6.41 \times 10^{-22}\ \text{V}\)
  2. \(10.0\ \text{V}\)
  3. \(1.00 \times 10^{16}\ \text{V}\)
  4. \(40.0\ \text{V}\)
Show Answers Only

\(D\)

Show Worked Solution

\(F=qE=\dfrac{qV}{d}\)

Rearrange the equation to make \(V\) the subject:

\(V=\dfrac{Fd}{q}=\dfrac{3.2 \times 10^{-15} \times 2 \times 10^{-3}}{1.602 \times 10^{-19}} = 40.0\ \text{V}\)

 \(\Rightarrow D\)

PHYSICS, M4 EQ-Bank 7

Two parallel conducting plates are separated by a distance of 25 mm and are connected to a 150 V DC power supply. An electron is placed in the region between the plates.

Calculate the magnitude of the force acting on the electron.   (2 marks)

Show Answers Only

\(9.6 \times 10^{-16}\ \text{N}\)

Show Worked Solution
\(F\) \(=qE\)  
  \(=\dfrac{qV}{d}\)  
  \(=\dfrac{1.602 \times 10^{-19} \times 150}{25 \times 10^{-3}}\)  
  \(= 9.6 \times 10^{-16}\ \text{N}\)  

PHYSICS, M4 EQ-Bank 4 MC

Two non-conducting spheres, \(\text{P}\) and \(\text{Q}\), are suspended from a horizontal insulating bar by identical length strings. Each sphere carries an electric charge and the strings are shown to be at different angles, as illustrated below. Assume the system is in static equilibrium and the diagram is to scale.
 

Which of the following pairs of statements best explains the behaviour of the system?

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \textbf{A} \rule[-1ex]{0pt}{0pt} & \text{• Sphere P has a greater charge than sphere Q.} \\ & \text{• The force of tension in P’s string is smaller than that in Q’s.} \\
\hline
\rule{0pt}{2.5ex} \textbf{B} \rule[-1ex]{0pt}{0pt} & \text{• The horizontal (electrostatic) force on P is equal in magnitude and opposite in direction to that on Q.} \\ & \text{• Sphere P is more massive than sphere Q.} \\
\hline
\rule{0pt}{2.5ex} \textbf{C} \rule[-1ex]{0pt}{0pt} & \text{• The net force acting on Q is greater than that on P.} \\ & \text{• The angle of Q's string is larger because it carries more charge.} \\
\hline
\rule{0pt}{2.5ex} \textbf{D} \rule[-1ex]{0pt}{0pt} & \text{• The larger angle on P's string means it experiences a stronger repulsive force.} \\ & \text{• This implies that the charge on P and Q are of opposite sign.} \\
\hline
\end{array}

Show Answers Only

\(B\)

Show Worked Solution
  • The electrostatic force between two charges is governed by Coulomb’s Law. Even if the spheres differ in mass or charge, the force each exerts on the other is always equal in magnitude and opposite in direction.
  • The strings are of equal length, but the angles are different, meaning one sphere is deflected more from the vertical. Both spheres are experiencing the same horizontal electrostatic force (magnitude), but if their angles differ, the vertical (gravitational) forces must be different to maintain equilibrium. This implies a difference in mass.
  • The greater the mass, the more vertical the string needs to be to balance the same horizontal force (smaller angle). So, the sphere with the smaller angle (less deflection) has a greater gravitational force pulling it down — hence, greater mass. Thus, \(\text{P}\) is more massive than \(\text{Q}\).

\(\Rightarrow B\)

PHYSICS, M4 EQ-Bank 6

  1. The diagram below shows a pair of parallel conducting plates. Using appropriate field conventions, draw electric field lines between the plates to illustrate the direction and uniformity of the electric field in the region between them.   (2 marks)
     

  

  1. The plates are separated by a distance of 8.0 cm, and a potential difference of 25 V is applied across them. Calculate the magnitude of the electric field in the region between the plates.   (1 mark)
  1. A particle of mass \(3.2 \times 10^{-20}\) kg carrying a charge of \(+2.4 \times 10^{-18}\ \text{C}\) is placed in the electric field, at a point 3 cm above the lower plate.
      
         

    Determine the magnitude and direction of the acceleration experienced by the particle due to the field.   (3 marks)

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a.    
       
 
b.    \(312.5\ \text{Vm}^{-1}\)

c.    \(2.34 \times 10^4\ \text{ms}^{-2},\ \text{up the page}\)

Show Worked Solution

a.    
     

  • The electric field should be represented by vertical lines that are equally spaced and extend from the positively charged plate to the negatively charged plate, indicating a uniform field.

b.    \(E = \dfrac{V}{d} = \dfrac{25}{0.08} = 312.5\ \text{Vm}^{-1}\)
  

c.    \(F = qE = 2.4 \times 10^{-18} \times 312.5 = 7.5 \times 10^{-16}\ \text{N, up the page}\)

Calculate acceleration using  \(F=ma:\)

   \(a =\dfrac{F}{m} = \dfrac{7.5 \times 10^{-16}}{3.2 \times 10^{-20}} = 2.34 \times 10^4\ \text{ms}^{-2},\ \text{up the page}\)

PHYSICS, M4 EQ-Bank 5

A positive test charge of +4.0 \(\mu\)C is placed between two parallel plates with a uniform electric field of 45 N/C. These plates are separated by 0.5 m, and the charge moves a distance of 0.2 m from point A to point B.
 

  1. Determine the magnitude and direction of the force on the charge.   (2 marks)
  1. Find the potential difference between point A and point B.   (1 mark)
  1. How much work does the electric field do on the charge during this movement?   (1 mark)
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a.    \( 1.8 \times 10^{-4}\ \text{N}\ \text{to the right}\)

b.    \(9\ \text{V}\)

c.    \(3.6 \times 10^{-5}\ \text{J}\)

Show Worked Solution

a.    \(F=qE = 4.0 \times 10^{-6} \times 45 = 1.8 \times 10^{-4}\ \text{N}\ \text{to the right}\)
 

b.    \(V = Ed = 45 \times 0.2 = 9\ \text{V}\)
 

c.    \(W= qEd = 4 \times 10^{-6} \times 45 \times 0.2 = 3.6 \times 10^{-5}\ \text{J}\)

PHYSICS, M4 EQ-Bank 3 MC

The following diagram shows the electric field between two charges.
 

Which of the following statements is correct?

  1. Charge \(q_1\) is positive and has a greater magnitude than \(q_2\).
  2. Charge \(q_1\) is positive and has a smaller magnitude than \(q_2\).
  3. Charge \(q_1\) is negative and has a greater magnitude than \(q_2\).
  4. Charge \(q_1\) is negative and has a smaller magnitude than \(q_2\).
Show Answers Only

\(B\)

Show Worked Solution
  • The field lines of an electric field are given by the direction that a small positive charge would follow. Hence \(q_1\) must be a positive charge as field lines run from \(q_1\) to \(q_2\)
  • The density of the field lines surrounding a point charge indicate the magnitude of the charge, with more lines corresponding to a greater magnitude of charge.
  • As there are more field lines around \(q_2\), it must be the stronger charge.

\(\Rightarrow B\)

PHYSICS, M4 EQ-Bank 4

Two small identical spheres \(\text{A}\) and \(\text{B}\) are suspended from a common point \(\text{Q}\) by light, non-conducting threads of equal length. Each sphere is given the same positive charge, and they repel each other, settling into equilibrium as shown.
 

The mass of each sphere is 3.0 g, and at equilibrium the horizontal separation between the spheres is 8.0 cm, while the length of each thread is 25 cm.

Calculate the magnitude of the charge on each sphere.   (4 marks)

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The magnitude of the charge on each sphere is  \(5.8 \times 10^{-8}\ \text{C}\).

Show Worked Solution
\(\sin\theta\) \(=\dfrac{4}{25}\)  
\(\theta\) \(=\sin^{-1}\left(\dfrac{4}{25}\right) = 9.21^{\circ}\)  

 
The following force diagram can be constructed from the diagram:

As the system is in equilibrium the weight force will be equal to \(T_v\)

\(T_v=mg=3 \times 10^{-3} \times 9.8=0.0294\ \text{N}\)

\(\tan(80.79)\) \(=\dfrac{T_v}{T_h}\)  
\(T_h\) \(=\dfrac{0.0294}{\tan80.79}=0.00477\ \text{N}\)  

 
As the system is in equilibrium, \(T_h = F_E\), therefore \(F_E = 0.00477\ \text{N}\)

\(F\) \(=\dfrac{1}{4 \pi \epsilon_0}\dfrac{q_1q_2}{r^2}\)  
\(0.00477\) \(=\dfrac{1}{4 \pi \epsilon_0}\dfrac{q^2}{0.08^2}\)  
\(q^2\) \(=3.397 \times 10^{-15}\)  
\(q\) \(=5.8 \times 10^{-8}\ \text{C}\)  

PHYSICS, M4 EQ-Bank 2 MC

During a dust storm on Mars, a research drone hovers between layers of charged dust clouds. The lower dust layer is at a potential of −80 MV relative to the upper layer, and the layers are 250 m apart. A tiny sensor on the drone accumulates a static charge of  \(q = 6 \times 10^{-12}\text{C.}\) 

What is the magnitude of the electric force acting on the charged sensor due to the electric field between the dust layers?

  1. \(1.92 \times 10^{-6}\ \text{N}\)
  2. \(3.84 \times 10^{-6}\ \text{N}\)
  3. \(5.76 \times 10^{-6}\ \text{N}\)
  4. \(1.28 \times 10^{-7}\ \text{N}\)
Show Answers Only

\(A\)

Show Worked Solution
\(F\) \(=\dfrac{qV}{d}\)  
  \(=\dfrac{6 \times 10^{-12} \times 80 \times 10^6}{250}\)  
  \(= 1.92 \times 10^{-6}\ \text{N}\)  

 
\(\Rightarrow A\)

PHYSICS, M4 EQ-Bank 3

Two small identical conducting spheres are mounted on insulating stands, 50 cm apart. One sphere carries a charge \((q_1)\) of +0.5 nC, and the other carries a charge \((q_2)\) of −0.2 nC.

  1. Calculate the magnitude and direction of the electrostatic force between the two spheres at this distance. Use Coulomb’s law and clearly state whether the force is attractive or repulsive.   (3 marks)
  1. The spheres are then briefly brought into contact, allowing charge to redistribute equally, and then separated again to a distance of 25 cm. What is the charge on each sphere now?   (1 mark)
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a.    \(3.60 \times 10^{-9}\ \text{N}\) towards each other.

b.    \(0.15\ \text{nC}\)

Show Worked Solution
a.    \(F\) \(=\dfrac{1}{4\pi\epsilon_0} \times \dfrac{q_1q_2}{r^2}\)
    \(=\dfrac{1}{4 \pi \times 8.854 \times 10^{-12}} \times \dfrac{0.5 \times 10^{-9} \times 0.2 \times 10^{-9}}{0.5^2}\)
    \(= 3.60 \times 10^{-9}\ \text{N}\ \ \text{towards each other}\)

 
b.    After contact, charges redistribute equally because:

  • Conductors allow free movement of charge.
  • Identical spheres have equal capacity to hold charge.
  • They reach the same electric potential when touched.
  •    \(q_{\text{each}} = \dfrac{q_1q_2}{2} = \dfrac{0.5 +(-0.2)}{2} = 0.15\ \text{nC}\).

PHYSICS, M4 EQ-Bank 2

Two parallel metal plates are 2.0 cm apart and are connected to a 300 V DC power supply. An alpha particle (a helium nucleus, charge = +2, mass = \(6.64 \times 10^{-27}\)) is released from rest at the positive plate.
 

  1. Calculate the magnitude and direction of the electric field between the plates.   (2 marks)
  1. Determine the final speed of the alpha particle as it reaches the negatively charged plate. Show all working.   (3 marks)
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a.    \(15000\ \text{Vm}^{-1}\) towards plate B.

b.    \(1.7 \times 10^{5}\ \text{ms}^{-1}\).

Show Worked Solution

a.    \(E=\dfrac{V}{d} = \dfrac{300}{0.02} = 15\,000\ \text{Vm}^{-1}\) towards plate B.
 

b.    Using  \(F=ma = qE\):

\(a = \dfrac{qE}{m} = \dfrac{2 \times 1.602 \times 10^{-19} \times 15\,000}{6.64 \times 10^{-27}} = 7.238 \times 10^{11}\ \text{ms}^{-2}\).

 
To determine the final speed of the alpha particle:

\(v^2\) \(=u^2 + 2as\)  
\(v\) \(= \sqrt{0^2 + 2 \times 7.238 \times 10^{11} \times 0.02}\)  
  \(= 1.7 \times 10^{5}\ \text{ms}^{-1}\).  

PHYSICS, M4 2023 VCE 1

Some physics students are conducting an experiment investigating both electrostatic and gravitational forces. They suspend two equally charged balls, each of mass 4.0 g, from light, non-conducting strings suspended from a low ceiling.

The charged balls repel each other with the strings at an angle of 60°, as shown in Figure 1.
 

There are three forces acting on each ball:

    • a tension force, \(T\)
    • a gravitational force, \(F_{g}\)
    • an electrostatic force, \(F_{E}\).
  1. On Figure 1, using the labels \(T, F_{g}\) and \(F_{E}\), draw each of the three forces acting on each of the charged balls.  (3 marks)

  2. Show that the tension force, \(T\), in each string is \(4.5 \times 10^{-2} \text{ N}\). Use  \(g=9.8 \text{ N kg}^{-1}\).
  3. Show your working.  (2 marks)

  4. Calculate the magnitude of the electrostatic force, \(F_{ E }\). Show your working.  (2 marks)

Show Answers Only

a.   
         

 
b.    System is in an equilibrium state:

  • The sum of the forces must add to zero as seen in the triangle below.
      

\(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\)

\(\sin\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{T}\)  
\(T\) \(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}=4.5 \times 10^{-2}\ \text{N}\)  

 
c.   Using the same triangle as part (b): 

\(\tan\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{F_E}\)  
\(F_E\) \(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}=2.3 \times 10^{-2}\ \text{N}\)  

Show Worked Solution

a.   
       

 
b.    System is in an equilibrium state:

  • The sum of the forces must add to zero as seen in the triangle below.
      

\(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\)

\(\sin\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{T}\)  
\(T\) \(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}=4.5 \times 10^{-2}\ \text{N}\)  

c.   Using the same triangle as part (b): 

\(\tan\,60^{\circ}\) \(=\dfrac{3.92 \times 10^{-2}}{F_E}\)  
\(F_E\) \(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}=2.3 \times 10^{-2}\ \text{N}\)  
♦ Mean mark 41%.

PHYSICS, M4 2023 VCE 2 MC

The diagram below shows two charges, \(Q_1\) and \(Q_2\), separated by a distance, \(d\).
 

 

There is a force, \(F\), acting between the two charges.

Which one of the following is closest to the magnitude of the force acting between the two charges if both \(d\) and the charge on \(Q_1\) are halved?

  1. \(\dfrac{F}{4}\)
  2. \(F\)
  3. \(2 F\)
  4. \(4 F\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Initially,}\ \ F=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{Q_1Q_2}{d^2}\)

 \(\text{After}\ \ Q_1\ \ \Rightarrow \ \ \dfrac{Q_1}{2}, \ \text{and}\ \ d\ \ \Rightarrow \ \ \dfrac{d}{2}:\)

\(F_{\text{new}}\) \(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{\frac{Q_1}{2}Q_2}{(\frac{d}{2})^2}\)  
  \(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{\frac{Q_1}{2}Q_2}{\frac{d^2}{4}}\)  
  \(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{4Q_1Q_2}{2d^2}\)  
  \(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{2Q_1Q_2}{d^2}\)  
  \(=2F\)  

 
\(\Rightarrow C\)

♦ Mean mark 55%.

PHYSICS, M4 2019 VCE 3*

Three charges \((- \text{Q} ,+2 \text{Q} ,-2 \text{Q})\) are placed at the vertices of an isosceles triangle, as shown below.
 


 

Draw a force vector diagram to show the direction of the net force on the charge \(- \text{Q}\) ?   (2 marks)

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Show Worked Solution
  • \(- \text{Q}\) will be attracted to \(+2 \text{Q}\) charge and repelled from the \(-2 \text{Q}\) charge.
  • This results in a force vector diagram as seen below.

PHYSICS, M4 2021 VCE 5b

Figure 5 shows a stationary electron \(\left( \text{e} ^{-}\right)\) 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.
 

 

The switch is now closed.

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

Show Answers Only

\(F=5.34 \times 10^{-15}\ \text{N towards the bottom plate.}\)

Show Worked Solution
  • Determine the magnitude of the electric force on the electron:
  •    \(F=qE = \dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6.0 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)
  • The longer line of the battery cell represents the positive terminal, hence the bottom plate will be the positive plate and the electric force on the electron will be towards the bottom plate.
♦ Mean mark 53%.

PHYSICS, M4 2015 HSC 24b

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

Electrons leave the cathode and are accelerated towards the anode.

Calculate the force on an electron due to the electric field between the cathode and the anode.   (2 marks)

Show Answers Only

\(F=4 \times 10^{-14}\) N

Show Worked Solution

Using  \(F=qE\)  and  \(E=\dfrac{V}{d}\):

\(F\) \(=\dfrac{qV}{d}\)  
  \(=\dfrac{1.602 \times 10^{-19} \times 5000}{0.02}\)  
  \(=4 \times 10^{-14}\) N  

PHYSICS, M4 2017 HSC 8 MC

An electron is fired in a vacuum towards a screen. With no electric field being applied, the electron hits the screen at \(P\). A uniform electric field is turned on and another electron is fired towards the screen from the same location, at the same velocity, striking the screen at point \(Q\).

With the electric field still turned on, a proton is fired towards the screen from the same starting point as the electrons and with the same velocity.

At what point does the proton strike the screen?

  1. \(A\)
  2. \(B\)
  3. \(C\)
  4. \(D\)
Show Answers Only

\(C\)

Show Worked Solution
  • As the proton has an equal but opposite charge to the electron it will experience a force in the opposite direction. Hence it will hit the screen to the right of \(P\).
  • Due to the proton having a greater mass, it will be deflected by a smaller amount and so will hit the screen at \(C\).

\(\Rightarrow C\)

♦♦♦ Mean mark 25%.