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Date May 2021 Marks available 3 Reference code 21M.2.HL.TZ2.3
Level Higher level Paper Paper 2 Time zone 2
Command term 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 centre of the ball, still carrying a charge of 1.2 × 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.

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=σ2ε0.

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 × 10−6 C. Determine σ.

[2]
b.ii.

The thread breaks. Explain the initial subsequent motion of the ball.

[3]
c.

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

[3]
d.i.

Outline, without calculation, whether or not the electric potential at P is zero.

[2]
d.ii.

Markscheme

identifies units of σ as C m-2 

Cm2×Nm2C2 seen and reduced to N C-1 

 

Accept any analysis (eg dimensional) that yields answer correctly

a.

horizontal force F on ball =Tsin30 ✓

T=mgcos30 

F «=mgtan30 = 0.025× 9.8 ×tan30» = 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=0.141.2×10-6«=1.2×105» ✓

σ=«2×8.85×10-12×0.141.2×10-6»=2.1×10-6«C m-2» 


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

Award [2] for a bald correct answer.

 

b.ii.

horizontal/repulsive force and vertical force/pull of gravity act on the ball

so ball has constant acceleration/constant net force

motion is in a straight line

at 30° to vertical away from wall/along original line of thread 

c.

Q0.222=1.2×10-60.182 ✓

«+»1.8×10-6«C»

2sf


Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

 

d.i.

work must be done to move a «positive» charge from infinity to P «as both charges are positive»
OR
reference to both potentials positive and added
OR
identifies field as gradient of potential and with zero value

therefore, point P is at a positive / non-zero potential ✓


Award [0] for bald answer that P has non-zero potential

d.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.

Syllabus sections

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