NESA Physics Problem Solving

1 sample question with marking guides and sample answers · Avg. score: 58.7%

Q5
2020
QCAA
Paper 2
4 marks
Q5
4 marks

An electron is situated halfway between two nuclei that are separated from each other by a distance of 4.5×10−104.5 \times 10^{-10} m. The first nucleus contains two protons. The second nucleus contains three protons.

Calculate the magnitude of the overall electromagnetic force experienced by the electron. (N to 1 decimal place)

Reveal Answer

F2=14πεoQ1×qer12F_2 = \dfrac{1}{4\pi\varepsilon_o}\dfrac{Q_1\times q_e}{r_1^2}

=9×109×3(1.6×10−19)×1.6×10−19(2.25×10−10)2= 9\times 10^9\times \dfrac{3(1.6\times 10^{-19})\times 1.6\times 10^{-19}}{(2.25\times 10^{-10})^2}

=1.3×10−8 N= 1.3\times 10^{-8}\ \text{N}

F1=14πεoQ2×qer22F_1 = \dfrac{1}{4\pi\varepsilon_o}\dfrac{Q_2\times q_e}{r_2^2}

=9×109×2(1.6×10−19)×1.6×10−19(2.25×10−10)2= 9\times 10^9\times \dfrac{2(1.6\times 10^{-19})\times 1.6\times 10^{-19}}{(2.25\times 10^{-10})^2}

=9.0×10−9 N= 9.0\times 10^{-9}\ \text{N}

∣Fnet∣=∣F1∣−∣F2∣|F_{net}| = |F_1| - |F_2|

=1.4×10−8−9.1×10−9= 1.4\times 10^{-8} - 9.1\times 10^{-9}

Force = 4.0×10−9 N4.0\times 10^{-9}\ \text{N} (to 1 decimal place)

Marking Criteria
DescriptorMarks

Indicates an understanding of the physical scenario in relation to Coulomb’s law.

1

Provides pertinent mathematical operation/s correctly performed.

1

Determines the forces (or electric field strength) imposed by each nuclei on the electron.

1

Determines the correct net force.

1

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