How many NESA Physics questions cover Charged Particles, Conductors and Electric and Magnetic Fields?

Q: How many NESA Physics questions cover Charged Particles, Conductors and Electric and Magnetic Fields?

AusGrader has 63 NESA Physics questions on Charged Particles, Conductors and Electric and Magnetic Fields, all with instant AI grading and detailed marking feedback.

NESA (HSC) Physics — Charged Particles, Conductors and Electric and Magnetic Fields Questions & Answers | AusGrader

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, $v$ , enters the chamber. Its velocity remains unchanged as it travels through the chamber.

A second proton having a velocity, $2v$ , 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.

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.

Q5

2021

QCAA

Paper 2

4 marks

Q5

4 marks

An alpha particle with a charge of $+3.2 \times 10^{-19}$ 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)

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

The work done on an object is equal to the change in kinetic energy.
$E_k = \frac{1}{2}mv^2$
$7.68 \times 10^{-17} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2$
$v^2 = \frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}$
$v = \sqrt{\frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}}$
Velocity = $1.5 \times 10^5 \text{ m s}^{-1}$ (to 2 significant figures)

Marking Criteria

Descriptor	Marks
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

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.

$E_k = W = Vq = (1.60 \times 10^{-19})(4.00 \times 10^4) = 6.40 \times 10^{-15} \text{ J}$

$E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}$

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

Marking Criteria

Descriptor	Marks
Equates $E_k$ to work done on electrons: $E_k = W = Vq$	1
Calculates value of $E_k$	1
Rearranges $E_k$ formula for $v$ : $E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}$	1
Substitutes values: $v = \sqrt{\frac{2(6.40 \times 10^{-15})}{9.11 \times 10^{-31}}}$	1
Calculates answer: $v = 1.19 \times 10^8 \text{ m s}^{-1}$	1

Q15b

5 marks

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

$E_t = E_k + E_{\text{rest}} \Rightarrow E_k = E_t - E_{\text{rest}}$
Then:

\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 $\beta$ :

\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 $\beta$ into $\beta = \sqrt{1 - \frac{v^2}{c^2}}$

\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

Descriptor	Marks
Uses mass-energy equivalence and total energy for $E_k$	1
Substitutes values	1
Solves for $\beta$	1
Substitutes $\beta$ into $\beta = \sqrt{1 - \frac{v^2}{c^2}}$ and solves for $v$ in terms of $c$	1
Calculates answer	1

Q15c

3 marks

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

$\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100 = 6.25\%$

Marking Criteria

Descriptor	Marks
Correct numerator and correct denominator: $\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100$	2
Expresses as % difference: $6.25\%$	1

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.

NESA Physics Charged Particles, Conductors and Electric and Magnetic Fields

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