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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%.