DP Physics (first assessment 2025)

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Question 21N.2.HL.TZ0.3

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Date November 2021 Marks available [Maximum mark: 11] Reference code 21N.2.HL.TZ0.3
Level HL Paper 2 Time zone TZ0
Command term Calculate, Determine, Draw, Explain, Show that, State Question number 3 Adapted from N/A
3.
[Maximum mark: 11]
21N.2.HL.TZ0.3

Two equal positive fixed point charges Q = +44 μC and point P are at the vertices of an equilateral triangle of side 0.48 m.

Point P is now moved closer to the charges.

A point charge q = −2.0 μC and mass 0.25 kg is placed at P. When x is small compared to d, the magnitude of the net force on q is F ≈ 115x.

An uncharged parallel plate capacitor C is connected to a cell of emf 12 V, a resistor R and another resistor of resistance 20 MΩ.

(a.i)

Show that the magnitude of the resultant electric field at P is 3 MN C−1

[2]

Markscheme

«electric field at P from one charge is kQr2=» 8.99×109×44×10-60.482

OR

1.7168×106 «NC−1» ✓


« net field is » 2×1.7168×106×cos30°=2.97×106 «NC−1» ✓

(a.ii)

State the direction of the resultant electric field at P.

[1]

Markscheme

directed vertically up «on plane of the page» ✓

 

Allow an arrow pointing up on the diagram.

(b.i)

Explain why q will perform simple harmonic oscillations when it is released.

[2]

Markscheme

force «on q» is proportional to the displacement ✓

and opposite to the displacement / directed towards equilibrium ✓

(b.ii)

Calculate the period of oscillations of q.

[2]

Markscheme

«a=Fm=»ω2x=115x0.25 ✓

T=«2πω=» 0.29«s» ✓

 

Award [2] marks for a bald correct answer.

Allow ECF for MP2.

(c.i)

At t = 0, the switch is connected to X. On the axes, draw a sketch graph to show the variation with time of the voltage VR across R.

[2]

Markscheme

decreasing from 12 ✓

correct shape as shown ✓

 

Do not penalize if the graph does not touch the t axis.

(c.ii)

The switch is then connected to Y and C discharges through the 20 MΩ resistor. The voltage Vc drops to 50 % of its initial value in 5.0 s. Determine the capacitance of C.

[2]

Markscheme

12=e-5.020×106 C ✓

C=3.6×10-7 «F» ✓

 

Award [2] for a bald correct answer.