NESA Physics Charged Particles, Conductors and Electric and Magnetic Fields

7 sample questions with marking guides and sample answers · Avg. score: 58.7%

Q5
2021
QCAA
Paper 2
4 marks
Q5
4 marks

An alpha particle with a charge of +3.2×10−19+3.2 \times 10^{-19} C moves through an electric field, accelerating from rest through a potential difference of 240 V.

Determine the velocity of the particle at the end of its acceleration, expressing your answer in scientific notation. (m/s to 2 significant figures)

Reveal Answer

The change in potential energy of an electric charge moving through an electric field is equivalent to the work done on the charge.
V=ΔUqV = \frac{\Delta U}{q}
ΔU=Vq\Delta U = Vq
=240×3.2×10−19= 240 \times 3.2 \times 10^{-19}
=7.68×10−17J=W= 7.68 \times 10^{-17} J = W

The work done on an object is equal to the change in kinetic energy.
Ek=12mv2E_k = \frac{1}{2}mv^2
7.68×10−17=12×6.64×10−27×v27.68 \times 10^{-17} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2
v2=7.68×10−1712×6.64×10−27v^2 = \frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}
v=7.68×10−1712×6.64×10−27v = \sqrt{\frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}}
Velocity = 1.5×105 m s−11.5 \times 10^5 \text{ m s}^{-1} (to 2 significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to work done on a moving charge in an electric field

1

Identifies that work done on the charge equates to its kinetic energy

1

Provides appropriate mathematical reasoning

1

Determines the velocity

1
Q14
2021
QCAA
Paper 1
1 mark
Q14
1 mark

Moving electric charges in a magnetic field experience

A

a decrease in charge.

B

an increase in charge.

C

a force parallel to the direction of the magnetic field.

D

a force perpendicular to the direction of the magnetic field.

Reveal Answer
A

a decrease in charge.

Electric charge is a fundamental conserved property and does not decrease due to motion or the presence of a magnetic field.

B

an increase in charge.

The magnitude of an electric charge is invariant; it does not increase regardless of its speed or the external fields applied.

C

a force parallel to the direction of the magnetic field.

The magnetic force is the result of a cross product (vecF=qvecvtimesvecB\\vec{F} = q\\vec{v} \\times \\vec{B}), which produces a vector perpendicular to the magnetic field, not parallel to it.

D

a force perpendicular to the direction of the magnetic field.

Correct Answer

According to the Lorentz force law, the magnetic force exerted on a moving charge is always perpendicular to both the velocity of the charge and the direction of the magnetic field.

Q19
2024
NESA
1 mark
Q19
1 mark

In a vacuum chamber there is a uniform electric field and a uniform magnetic field.

A proton having a velocity, vv, enters the chamber. Its velocity remains unchanged as it travels through the chamber.

A second proton having a velocity, 2v2v, in the same direction as the first proton, then enters the chamber at the same point as the first proton.

In the chamber, the acceleration of the second proton

A

is zero.

B

is constant in magnitude and direction.

C

changes in both magnitude and direction.

D

is constant in magnitude, but not direction.

Reveal Answer
A

is zero.

The magnetic force depends on velocity (FB=qvBF_B = qvB), so doubling the velocity doubles the magnetic force. This unbalances it with the constant electric force, resulting in a non-zero net force and acceleration.

B

is constant in magnitude and direction.

The initial net force changes the proton's velocity vector. Since the magnetic force depends on this changing velocity, the net force and acceleration cannot remain constant.

C

changes in both magnitude and direction.

Correct Answer

The unbalanced forces cause the proton's velocity to change. Because the magnetic force depends on the instantaneous velocity vector (qv⃗×B⃗q\vec{v} \times \vec{B}), the net force and resulting acceleration will continuously change in both magnitude and direction.

D

is constant in magnitude, but not direction.

As the proton's velocity changes in both magnitude and direction relative to the magnetic field, the magnitude of the magnetic force (and thus the net acceleration) will also change, not just its direction.

Q10
2022
QCAA
Paper 1
1 mark
Q10
1 mark

Electric field strength refers to the

A

intensity of an electric field at a particular location.

B

change in electrical potential energy between two defined points.

C

sum of electrically charged particles passing a point in a given time.

D

physical property of an object experiencing a force in an electromagnetic field.

Reveal Answer
A

intensity of an electric field at a particular location.

Correct Answer

Electric field strength is a measure of the intensity or magnitude of the electric field at a specific point, defined as the force exerted per unit positive charge (E=F/qE = F/q).

B

change in electrical potential energy between two defined points.

The change in electrical potential energy per unit charge between two points defines electric potential difference (voltage), not electric field strength.

C

sum of electrically charged particles passing a point in a given time.

The rate at which charged particles pass a specific point defines electric current (I=Q/tI = Q/t), not electric field strength.

D

physical property of an object experiencing a force in an electromagnetic field.

This describes electric charge, which is the property of matter that causes it to experience a force, whereas electric field strength describes the field itself.

Q17
2021
QCAA
Paper 1
1 mark
Q17
1 mark

Electrical potential energy is the

A

intensity of an electric field at a particular location.

B

difference in potential that tends to give rise to an electric current.

C

capacity of electric charge carriers to do work due to their position in an electric circuit.

D

work done on an electron in accelerating it through an electrical potential difference of one volt.

Reveal Answer
A

intensity of an electric field at a particular location.

This describes electric field strength (EE), which is the force exerted per unit charge at a specific location, not the energy.

B

difference in potential that tends to give rise to an electric current.

This describes electric potential difference (voltage), which is the difference in electric potential energy per unit charge between two points.

C

capacity of electric charge carriers to do work due to their position in an electric circuit.

Correct Answer

Electrical potential energy is the stored energy a charge possesses due to its position in an electric field, representing the capacity to do work.

D

work done on an electron in accelerating it through an electrical potential difference of one volt.

This is the specific definition of an electron-volt (eV), a unit of energy, rather than the general definition of electrical potential energy.

Q19
2024
QCAA
Paper 1
1 mark
Q19
1 mark

A 7 μ\muC charge requires 1.5×10−81.5 \times 10^{-8} J of energy to be moved between two points in an electric field.

What is the order of magnitude of the potential difference between the two points?

A

10−210^{-2} V

B

10−310^{-3} V

C

10−910^{-9} V

D

10−1310^{-13} V

Reveal Answer
A

10−210^{-2} V

This is incorrect. While the calculation yields approximately 0.21×10−20.21 \times 10^{-2} V, proper scientific notation requires the coefficient to be between 1 and 10 (2.1×10−32.1 \times 10^{-3} V), making the order of magnitude 10−310^{-3}.

B

10−310^{-3} V

Correct Answer

Using the formula V=WqV = \frac{W}{q}, the potential difference is 1.5×10−8 J7×10−6 C≈2.1×10−3\frac{1.5 \times 10^{-8} \text{ J}}{7 \times 10^{-6} \text{ C}} \approx 2.1 \times 10^{-3} V. Since the coefficient 2.1 is less than 10≈3.16\sqrt{10} \approx 3.16, the order of magnitude is 10−310^{-3}.

C

10−910^{-9} V

This value is far too small. It likely results from an arithmetic error in handling the exponents during division.

D

10−1310^{-13} V

This incorrect answer results from multiplying the work and charge (W×qW \times q) instead of dividing them. The product would be approximately 1.05×10−131.05 \times 10^{-13}.

Q15
2025
SCSA
13 marks
Q15

Electrons are accelerated from rest across a potential difference of 40.0 kV.

Q15a
5 marks

Calculate the final speed of the electrons using Newtonian physics, which ignores relativistic effects.

Reveal Answer

Ek=W=Vq=(1.60×10−19)(4.00×104)=6.40×10−15 JE_k = W = Vq = (1.60 \times 10^{-19})(4.00 \times 10^4) = 6.40 \times 10^{-15} \text{ J}

Ek=12mv2⇒v=2EkmE_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}

Hence:
v=2(6.40×10−15)9.11×10−31=1.19×108 m s−1v = \sqrt{\frac{2(6.40 \times 10^{-15})}{9.11 \times 10^{-31}}} = 1.19 \times 10^8 \text{ m s}^{-1}

Marking Criteria
DescriptorMarks

Equates EkE_k to work done on electrons: Ek=W=VqE_k = W = Vq

1

Calculates value of EkE_k

1

Rearranges EkE_k formula for vv: Ek=12mv2⇒v=2EkmE_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}

1

Substitutes values: v=2(6.40×10−15)9.11×10−31v = \sqrt{\frac{2(6.40 \times 10^{-15})}{9.11 \times 10^{-31}}}

1

Calculates answer: v=1.19×108 m s−1v = 1.19 \times 10^8 \text{ m s}^{-1}

1
Q15b
5 marks

Calculate the final speed of the electrons using Einstein’s special theory of relativity.

Reveal Answer

Et=Ek+Erest⇒Ek=Et−ErestE_t = E_k + E_{\text{rest}} \Rightarrow E_k = E_t - E_{\text{rest}}
Then:

Ek=mc21−v2c2−mc2=(1β−1)mc2 where β=1−v2c2\begin{align*} E_k &= \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2\\ &= \left(\frac{1}{\beta} - 1\right)mc^2 \text{ where } \beta = \sqrt{1 - \frac{v^2}{c^2}} \end{align*}

Solve for β\beta:

6.40×10−15=(1β−1)(9.11×10−31)(3.00×108)21β=6.40×10−15(9.11×10−31)(3.00×108)2+11β=1.0781β=0.9276\begin{align*} 6.40 \times 10^{-15} &= \left(\frac{1}{\beta} - 1\right)(9.11 \times 10^{-31})(3.00 \times 10^8)^2\\ \frac{1}{\beta} &= \frac{6.40 \times 10^{-15}}{(9.11 \times 10^{-31})(3.00 \times 10^8)^2} + 1\\ \frac{1}{\beta} &= 1.0781\\ \beta &= 0.9276 \end{align*}

Sub β\beta into β=1−v2c2\beta = \sqrt{1 - \frac{v^2}{c^2}}

0.9276=1−v2c2β=1−v2c2 and solves for v in terms of cv2c2=1−(0.9276)2v=0.1396 cv=0.3736 cv=1.12×108 m s−1\begin{align*} 0.9276 &= \sqrt{1 - \frac{v^2}{c^2}}\\ \beta &= \sqrt{1 - \frac{v^2}{c^2}} \text{ and solves for } v \text{ in terms of } c\\ \frac{v^2}{c^2} &= 1 - (0.9276)^2\\ v &= \sqrt{0.1396}\,c\\ v &= 0.3736\,c\\ v &= 1.12 \times 10^8 \text{ m s}^{-1} \end{align*}
Marking Criteria
DescriptorMarks

Uses mass-energy equivalence and total energy for EkE_k

1

Substitutes values

1

Solves for β\beta

1

Substitutes β\beta into β=1−v2c2\beta = \sqrt{1 - \frac{v^2}{c^2}} and solves for vv in terms of cc

1

Calculates answer

1
Q15c
3 marks

Calculate the percentage difference of your answer to part (a) compared to part (b).

Reveal Answer

1.19×108−1.12×1081.12×108×100=6.25%\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100 = 6.25\%

Marking Criteria
DescriptorMarks

Correct numerator and correct denominator: 1.19×108−1.12×1081.12×108×100\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100

2

Expresses as % difference: 6.25%6.25\%

1

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