IB Questionbank

Updates to Questionbank

User interface language: English | Español

Date	May 2021	Marks available	2	Reference code	21M.2.SL.TZ2.3
Level	Standard level	Paper	Paper 2	Time zone	2
Command term	State and Calculate	Question number	3	Adapted from	N/A

Question

A vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.

The charge per unit area on the surface of the wall is σ. It can be shown that the electric field strength E due to the charge on the wall is given by the equation

$E = \frac{σ}{2 ε_{0}}$ .

Demonstrate that the units of the quantities in this equation are consistent.

[2]

a.

The thread makes an angle of 30° with the vertical wall. The ball has a mass of 0.025 kg.

Determine the horizontal force that acts on the ball.

[3]

b.i.

The charge on the ball is 1.2 × 10⁻⁶C. Determine σ.

[2]

b.ii.

The centre of the ball, still carrying a charge of $1.2 \times 10^{- 6} C$ , is now placed $0.40 m$ from a point charge Q. The charge on the ball acts as a point charge at the centre of the ball.

P is the point on the line joining the charges where the electric field strength is zero.
The distance PQ is $0.22 m$ .

Calculate the charge on Q. State your answer to an appropriate number of significant figures.

[2]

c.

Markscheme

identifies units of $σ$ as ${C m}^{- 2}$ ✓

$\frac{C}{m^{2}} \times \frac{N m^{2}}{C^{2}}$ seen and reduced to ${N C}^{- 1}$ ✓

Accept any analysis (eg dimensional) that yields answer correctly

a.

horizontal force $F$ on the ball $= T \sin 30$ ✓

$T = \frac{m g}{\cos 30}$ ✓

$F « = m g \tan 30 = 0.025 \times 9.8 \times \tan 30 » = 0.14 « N »$ ✓

Allow g = 10 N kg⁻¹

Award [3] marks for a bald correct answer.

Award [1max] for an answer of zero, interpreting that the horizontal force refers to the horizontal component of the net force.

b.i.

$E = \frac{0.14}{1.2 \times 10^{- 6}} « = 1.2 \times 10^{5} »$ ✓

$σ = « \frac{2 \times 8.85 \times 10^{- 12} \times 0.14}{1.2 \times 10^{- 6}} » = 2.1 \times 10^{- 6} « {C m}^{- 2} »$ ✓

Allow ECF from the calculated F in (b)(i)

Award [2] for a bald correct answer.

b.ii.

$\frac{Q}{0 . 22^{2}} = \frac{1.2 \times 10^{- 6}}{0 . 18^{2}}$ ✓

$« + » 1.8 \times 10^{- 6} « C »$ ✓

2sf ✓

Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

c.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.

Syllabus sections

Core » Topic 5: Electricity and magnetism » 5.1 – Electric fields

Show 60 related questions

Hide related questions

Core » Topic 1: Measurements and uncertainties » 1.1 – Measurements in physics

Core » Topic 1: Measurements and uncertainties

Core » Topic 5: Electricity and magnetism