DP Physics (first assessment 2025)

Question SPM.2.SL.TZ0.7

Date	May Specimen-2023	Marks available	[Maximum mark: 20]	Reference code	SPM.2.SL.TZ0.7
Level	SL	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Draw, Explain, Outline, Show that, State	Question number	7	Adapted from	N/A

[Maximum mark: 20]

SPM.2.SL.TZ0.7

The diagram shows two parallel conducting plates that are oppositely charged.

(a.i)

Draw the electric field lines due to the charged plates.

[2]

Markscheme

equally spaced arrows «by eye» all pointing down ✓

edge effects also shown with arrows ✓

(a.ii)

The potential difference between the plates is 960 V and the distance between them is 8.0 mm. Calculate the electric field strength E between the plates.

[2]

Markscheme

$E = \frac{V}{d} = \frac{960}{8.0 \times 10^{- 3}}$ ✓

$E = 1.2 \times 10^{5}$ «NC⁻¹» ✓

In an experiment, an oil drop is introduced into the space between the plates through a small hole in the upper plate. The oil drop moves through air in a tube before falling between the plates.

(b)

Explain why the oil drop becomes charged as it falls through the tube.

[1]

Markscheme

friction transfers electron(s) to or from drop

AND

through collisions/ interaction with air molecules in the tube OR through collisions/interaction with wall of tube ✓

(c)

The oil drop is observed to be stationary in the space between the plates. Buoyancy is one of the forces acting on the drop.

The density of oil is 730 times greater than that of air.

(c.ii)

Draw the forces acting on the oil drop, ignoring the buoyancy force.

[2]

Markscheme

Weight vertically down AND electric force vertically up ✓

Of equal length «by eye» ✓

(c.i)

Show that the buoyancy force is much smaller than the weight.

[3]

Markscheme

weight of oil drop is $ρ_{o} g V$ ✓

$\frac{F_{b}}{W} = \frac{ρ_{a} g V}{ρ_{o} g V} = \frac{ρ_{a}}{ρ_{o}}$ ✓

« $\frac{F_{b}}{W} = \frac{1}{730} =$ » $1.4 \times 10^{- 3}$

Ratio of $F_{b}$ to $W$ is much less than 1 ✓

(c.iii)

Show that the electric charge on the oil drop is given by

$q = \frac{ρ_{o} g V}{E}$

where $ρ_{o}$ is the density of oil and $V$ is the volume of the oil drop.

[2]

Markscheme

Mass of drop is $ρ_{0} V$ ✓

$q E = (ρ_{0} V) g$ ✓

«hence answer»

MP1 must be shown implicitly for credit.

(c.iv)

State the sign of the charge on the oil drop.

[1]

Markscheme

Negative ✓

(d)

The electric field is turned off. The oil drop falls vertically reaching a constant speed v.

(d.i)

Outline why, for this drop, $ρ_{o} g V = 6 π η r v$ where $η$ is the viscosity of air and $r$ is the radius of the oil drop.

[2]

Markscheme

Net force is zero ✓

Acceleration of the oil drop is zero ✓

For terminal velocity drag must equal weight ✓

weight $= ρ_{o} g V$ and drag $= 6 π η r v$ ✓

(d.ii)

Show that the charge on the oil drop is about $4.8 \times 10^{- 19} C$ .

The following data for the oil drop are available:

$r = 1.36 μm$
$η = 1.60 \times 10^{- 5} Pa s$
$v = 0.140 {mm s}^{- 1}$

[3]

Markscheme

$q = \frac{6 π η r v}{E}$ ✓

$q = \frac{6 π \times 1.60 \times 10^{- 5} \times 1.36 \times 10^{- 6} \times 1.40 \times 10^{- 4}}{1.2 \times 10^{5}}$ ✓

$q = 4.79 \times 10^{- 19}$ «C» ✓

Answer must be shown to 3+ sf.

(d.iii)

The oil drop splits into two parts of equal mass. Both are charged. Deduce the net charge on each part.

[2]

Markscheme

charge is quantized ✓

so, the charges must be 1e and 2e ✓

Syllabus sections

D. Fields

D. Fields » D.2 Electric and magnetic fields

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