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PHYSICS, M4 2017 VCE 1

Three charges are arranged in a line, as shown in Figure 1.
 

Draw an arrow at point \(\text{X}\) to show the direction of the resultant electric field at \(\text{X}\). If the resultant electric field is zero, write the letter ' \(\text{N}\) ' at \(\text{X}\).  (2 mark)

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(Insert image here)

Show Worked Solution

→ The electric field direction will be from the positive charge to the negative charge.

→The \(-Q\) on the right has no effect to the direction of the electric field at \(\text{X}\).

(Insert image here)

PHYSICS, M4 2017 VCE 3*

Two large charged plates with equal and opposite charges are placed close together, as shown in the diagram below.

A distance of 5.0 mm separates the plates. The electric field between the plates is equal to 1000 N C\(^{-1}\).
 

Calculate the voltage difference between the plates.   (2 marks)

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Voltage difference is \(5\ \text{V}\).

Show Worked Solution
\(V\) \(=Ed\)  
  \(=1000 \times 5.0 \times 10^{-3}\)  
  \(=5\ \text{V}\)  

PHYSICS, M4 2019 VCE 2

Figure 2 shows two equal positive stationary point charges placed near each other.

 

 
 

 

Sketch on Figure 2 the shape and direction of the electric field lines. Use at least eight field lines.  (2 marks) 

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→ Field lines must not touch.

→ Fields will repel one another.

PHYSICS, M4 2019 VCE 2*

The electric field between two parallel plates that are 1.0 × 10\(^{-2}\) m apart is 2.0 × 10\(^{-4}\) N C\(^{-1}\).

Calculate the voltage between the plates?   (2 marks)

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\(2.0 \times 10^{-6}\ \text{V} \)

Show Worked Solution
\(E\) \(=\dfrac{V}{d}\)  
\(V\) \(=E \times d\)  
  \(=2.0 \times 10^{-4} \times 1.0 \times 10^{-2}\)  
  \(=2.0 \times 10^{-6}\ \text{V} \)  

PHYSICS, M4 2020 VCE 1 MC

The diagram below shows the electric field lines between two charges of equal magnitude.
 

The best description of the two charges is that the

  1. charges are both positive.
  2. charges are both negative.
  3. charges can be either both positive or both negative.
  4. left-hand charge is positive and the right-hand charge is negative.
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\(A\)

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→ The field lines show the charges are experiencing a repulsion force hence they must be the same charge.

→ The direction of the field lines are away from the charges so they must be positive charges.

\(\Rightarrow A\)

PHYSICS, M4 2021 VCE 3 MC

The diagram below shows two parallel metal plates with opposite charges on each plate. \(\text{X , Y}\) and \(\text{Z}\) represent different distances from the positive plate.
 

Which one of the following graphs best shows the electric field strength, \(E\), versus the position, \(x\), between the two parallel plates?
 

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\(A\)

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→ The strength of the electric field is uniform between parallel plates. Hence its magnitude remains constant.

\( \Rightarrow A\)

PHYSICS, M4 2021 VCE 2 MC

The diagram below shows the electric field lines between four charged spheres: \(\text{P, Q, R}\) and \(\text{S}\). The magnitude of the charge on each sphere is the same.
  
 


 

Which of the following correctly identifies the type of charge (+ positive or – negative) that resides on each of the spheres \(\text{P, Q, R}\) and \(\text{S}\)?

  \(\textbf{P}\) \(\textbf{Q}\) \(\textbf{R}\) \(\textbf{S}\)
A.   \(\quad - \quad\) \(\quad + \quad\) \(\quad - \quad\) \(\quad + \quad\)
B. \(\quad + \quad\) \(\quad - \quad\) \(\quad + \quad\) \(\quad - \quad\)
C. \(\quad - \quad\) \(\quad - \quad\) \(\quad + \quad\) \(\quad + \quad\)
D. \(\quad + \quad\) \(\quad + \quad\) \(\quad - \quad\) \(\quad - \quad\)
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\(B\)

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→ Electric field lines travel away from positive charges and towards negative charges.

→ Therefore, \(\text{P}\) and \(\text{R}\) must be positive charges.

\( \Rightarrow B\)

PHYSICS, M4 2022 VCE 4*

Two point charges, \(Q\) and \(4Q\), are placed 12 cm apart, as shown in the diagram below.
 

On the straight line between the charges \(Q\) and \(4Q\), find where the electric field is zero.   (3 marks)

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\(\text{4cm to the right of}\ Q\)

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→ The point where the force on a charged particle \(q\) as a result of the two fields from \(Q\) and \(4Q\) is equal and opposite is where the electric field strength would be zero.

→ Let \(x\) equal the distance from \(Q\) to \(q\) and \(y\) be the distance from \(4Q\) to \(q\) so  \(x +y=12\)
 

\(F_{\text{\(Q\) on \(q\)}}\) \(=F_{\text{\(4Q\) on \(q\)}}\)  
\(\dfrac{1}{4 \pi \varepsilon_0} \dfrac{Q \times q}{x^2}\) \(=\dfrac{1}{4 \pi \varepsilon_0} \dfrac{4Q \times q}{y^2}\)  
\(\dfrac{1}{x^2}\) \(=\dfrac{4}{y^2}\)  
\(y^2\) \(=4x^2\)  
\(y\) \(=2x\)  

 

→ \(x+2x=12\ \ \Rightarrow\ \ x=4\ \text{cm} \)

♦♦♦ Mean mark 18%.

PHYSICS, M4 2012 HSC 6 MC

The diagram represents the electric field around a negative charge.
 

   

If the magnitude of the charge were doubled, which diagram would best represent the new electric field?

 


 

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\(C\)

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→ The density of the field lines corresponds to the strength of the electric field surrounding the point charge.

→ Hence a stronger charge produces a stronger field and so the field lines become denser.

\(\Rightarrow C\)

PHYSICS, M4 2013 HSC 26a

An electric field is produced between two charged parallel plates, \(M\) and \(N\).
 

The plates, \(M\) and \(N\), are 1.0 cm apart and have an electric field of 15 V m\(^{-1}\).

Calculate the potential difference between the plates.  (2 marks)

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\(0.15 \text{ V}\) 

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\(\text{Convert units:  1 cm = 0.01 m}\).

\(E\) \(=\dfrac{V}{d}\)  
\(V\) \(=Ed\)  
  \(=15 \times 0.01\)  
  \(=0.15 \text{ V}\)  

PHYSICS, M4 2013 HSC 14 MC

Two charged plates are initially separated by a distance as shown in the diagram.
 

The potential difference between the plates remains constant.

Which of the graphs best represents the change in electric field strength as the distance between the two plates is increased?
 


 

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\(B\)
Show Worked Solution

→The electric field strength and distance vary according to the equation:

\(E=\dfrac{V}{d}\)

→ Thus \(E\) and \(d\) are inversely proportionally which is depicted in \(B\).

\(\Rightarrow B\)

PHYSICS, M4 2017 HSC 3 MC

Which of the following correctly shows the electric field between two parallel, charged plates?
 


 

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\(A\)

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→ The electric field lines are defined by the direction of force on a positive charge.

→ A positive charge will experience a repulsive force from the positive plate and attractive force to the negative plate.

\(\Rightarrow A\)

PHYSICS, M6 2016 HSC 5 MC

The diagram shows two parallel charged plates `5 × 10^(-3)  text{m}` apart.
 

What is the magnitude of the electric field between the plates in `text{V m}^(-1)` ?

  1. `3.3 × 10^(-4)`
  2. `0.33`
  3. `3`
  4. `3000`
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`D`

Show Worked Solution
`E` `=(V)/(d)`  
  `=(15)/(0.005)`  
  `=3000\ text{V m}^(-1)`  

 
`=>D`

PHYSICS, M6 2018 HSC 26

Outline the similarities and differences between the effects of electric fields and gravitational fields on matter. In your answer, refer to the definitions of these fields.   (4 marks)

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Differences in effects:

→ Electric fields are regions in space where a force acts on a charged particle whereas gravitational fields are regions in space where a force acts on an object with mass due to the influence of another object with mass.

→ Electric field strength is defined by `E=(F)/(q)`, or force per unit charge.

→ Gravitational field strength is defined with `g=(F)/(m)`, or force per unit mass.

→ Electric fields can produce forces of repulsion (due to like charges) while gravitational fields can only produce forces of attraction.
 

Similarities:

→ Both fields are able to produce forces of attraction.

→ Electric and gravitational fields vary in strength due to magnitudes of charge and mass respectively.

→ The strength of both fields is inversely proportional to the distance of separation from a point charge or mass.

→ Gravity is a significantly weaker fundamental force compared to electromagnetism.

Show Worked Solution

Differences in effects:

→ Electric fields are regions in space where a force acts on a charged particle whereas gravitational fields are regions in space where a force acts on an object with mass due to the influence of another object with mass.

→ Electric field strength is defined by `E=(F)/(q)`, or force per unit charge.

→ Gravitational field strength is defined with `g=(F)/(m)`, or force per unit mass.

→ Electric fields can produce forces of repulsion (due to like charges) while gravitational fields can only produce forces of attraction.
 

Similarities:

→ Both fields are able to produce forces of attraction.

→ Electric and gravitational fields vary in strength due to magnitudes of charge and mass respectively.

→ The strength of both fields is inversely proportional to the distance of separation from a point charge or mass.

→ Gravity is a significantly weaker fundamental force compared to electromagnetism.