MYP · 13 · extended response
An electrostatic sprayer and an unintended spark
Electricity · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Nozzle charge versus potential
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Charge / μC
Horizontal axis (x): Potential / kV
Read the exact data as a table
| Potential / kV | Charge / μC |
|---|---|
| 0 | 0 |
| 10 | 1 |
| 20 | 2 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A conducting sprayer nozzle behaves as a 100 pF capacitor relative to its surroundings. The supplied charge–potential data describe it. It operates at +20 kV and transfers 2.0 nC to each paint droplet. Assume identical droplets. A grounding connection can discharge the nozzle; no calculation here establishes a safe ignition threshold. Supplied model: Use Q = CV and stored electrostatic energy U = ½QV. Supplied data (Potential / kV; Charge / μC): (0, 0), (10, 1), (20, 2).
- (a)
Find charge stored, stored energy and the number of charged droplets corresponding to that stored charge.
3 marks - (b)
Explain why charged paint is attracted to a grounded conductor and how grounding differs from insulation as risk control.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Find charge stored, stored energy and the number of charged droplets corresponding to that stored charge.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The graph has C = Q/V = (1.0×10⁻⁶)/(1.0×10⁴) = 100 pF; at 20 kV charge is 2.0×10⁻⁶ C.
- 2
Stored energy is ½QV = ½(2.0×10⁻⁶)(20000) = 0.020 J. The factor ½ accounts for the potential rising during charging.
- 3
Q/qdrop = (2.0×10⁻⁶)/(2.0×10⁻⁹) = 1000 droplets if that charge were transferred with no replenishing supply.
(b)
Explain why charged paint is attracted to a grounded conductor and how grounding differs from insulation as risk control.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
A positive droplet induces negative surface charge on the nearby conducting workpiece. Grounding lets electrons flow from Earth, maintaining a suitable potential and sustained attraction.
- 2
Conductors allow mobile charge to redistribute; an insulator can retain local deposited charge. Static charge is not necessarily a continuous electric current.
- 3
Bonding and grounding reduce charge accumulation and potential differences. Insulation can prevent contact current but may retain static charge, so its role must match the actual discharge or ignition risk.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- The graph has C = Q/V = (1.0×10⁻⁶)/(1.0×10⁴) = 100 pF; at 20 kV charge is 2.0×10⁻⁶ C.
- Stored energy is ½QV = ½(2.0×10⁻⁶)(20000) = 0.020 J. The factor ½ accounts for the potential rising during charging.
- Q/qdrop = (2.0×10⁻⁶)/(2.0×10⁻⁹) = 1000 droplets if that charge were transferred with no replenishing supply.
- A positive droplet induces negative surface charge on the nearby conducting workpiece. Grounding lets electrons flow from Earth, maintaining a suitable potential and sustained attraction.
- Conductors allow mobile charge to redistribute; an insulator can retain local deposited charge. Static charge is not necessarily a continuous electric current.
- Bonding and grounding reduce charge accumulation and potential differences. Insulation can prevent contact current but may retain static charge, so its role must match the actual discharge or ignition risk.
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