MYP Physics · topic MYP-13.01
Static charge, conductors, and insulators
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
An electrostatic sprayer and an unintended spark
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]
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.
Show a hint
Begin with part (a): Find charge stored, stored energy and the number of charged droplets corresponding to that stored charge.
Reveal the worked solution
- 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.
Answer (a) Q = 2.0 μC; stored energy = 0.020 J; equivalent droplet count = 1000. (b) Charge redistribution produces attraction; grounding provides a discharge path while insulation restricts charge flow.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.