IBDP Physics 5.1 – Electric fields : IB style Question Bank – SL Paper 1

IB PHYSICS SL (Standard level)- 2024 – Practice Questions- All Topics

Topic 5.1 – Electric Fields

Topic 5 Weightage : 8 % 

All Questions for Topic 5.1 – Charge , Electric field , Coulomb’s law , Electric current , Direct current (dc) , Potential difference

Question

Two charges Q_1 and Q_2, each equal to 2 nC, are separated by a distance 3 m in a vacuum.

What is the electric force on Q_2 and the electric field due to Q_1 at the position of Q_2?

▶️Answer/Explanation

Ans: A

\(F_{Q_2}=k\frac{Q_1Q_2}{r^2}\)
\(=\frac{9\times 10^9 \times 2\times 10^{-9}\times 2\times 10^{-9}}{3^2}=4\times 10^{-9}\)
\(E_{Q_{21}}=\frac{F}{Q_1}=\frac{4\times 10^{-9}}{2\times 10^{-9}}=2\; NC^{-1}\)

Question

A metal wire has n free charge carriers per unit volume. The charge on the carrier is q. What additional quantity is needed to determine the current per unit area in the wire?

A Cross-sectional area of the wire

B Drift speed of charge carriers

C Potential difference across the wire

D Resistivity of the metal

▶️Answer/Explanation

Ans: B

The uniform current I, flowing through the conductor is given by
I = n e A vd
where n = number of free electrons per unit volume,
A = area of cross-section,  vd = drift velocity
In vector form, \(\Rightarrow \vec{j} =-ne \vec{v_d}\)
The negative sign is because the direction of drift velocity of electron is opposite to \(\vec{j}\)

Question

An electron is accelerated through a potential difference of 2.5 MV. What is the change in kinetic energy of the electron?

A.  0.4μJ

B.  0.4 nJ

C.  0.4 pJ

D.  0.4 fJ

▶️Answer/Explanation

Markscheme

C

work done = change in kinetic energy 

\(=qV = e V= 1.6\times 10^{-19}\times 2.5 \times 10^6 = 4 \times 10^{-13} =0.4 \times 10^{-12} =0.4 \; pJ\)

Question

Which of the following diagrams illustrates the electric field pattern of a negatively charged sphere?

▶️Answer/Explanation

Markscheme

A

The lines of force diverge out from a positive charge and converge at a negative charge. i.e. the lines of force are always directed from higher to lower potential.
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