MYP · 15 · extended response
Finding the location where a field vanishes
Fields and gravity · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 3
- Answer
- Complete
Field of the left charge alone
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Field magnitude / N C⁻¹
Horizontal axis (x): Distance from left charge / m
Read the exact data as a table
| Distance from left charge / m | Field magnitude / N C⁻¹ |
|---|---|
| 0.1 | 3600 |
| 0.2 | 900 |
| 0.3 | 400 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. Two fixed point charges lie on the x-axis: +4.0 nC at x = 0 and +1.0 nC at x = 0.30 m. Use k = 9.0×10⁹ N m² C⁻² and V = kQ/r with zero potential at infinity. The supplied graph gives the field magnitude of the +4.0 nC charge alone. Supplied model: For point charges use field magnitude E = k|Q|/r² and force F = qE, adding field vectors with directions. Supplied data (Distance from left charge / m; Field magnitude / N C⁻¹): (0.1, 3600), (0.2, 900), (0.3, 400).
- (a)
Locate the zero-resultant-field point between the charges and find the electric potential there.
3 marks - (b)
Find force on a −2.0 nC test charge at x = 0.10 m, and describe useful field-line features without treating them as literal objects.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Locate the zero-resultant-field point between the charges and find the electric potential there.
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
Between two positive charges their fields oppose. Equality requires 4/x² = 1/(0.30−x)², giving x/(0.30−x) = 2.
- 2
Solve x = 0.20 m, with distances 0.20 and 0.10 m. Field directions, not just magnitudes, must be checked.
- 3
Potential adds as a scalar: V = 9.0×10⁹[(4.0×10⁻⁹)/0.20+(1.0×10⁻⁹)/0.10] = 270 V. Zero field does not require zero potential.
(b)
Find force on a −2.0 nC test charge at x = 0.10 m, and describe useful field-line features without treating them as literal objects.
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
At x = 0.10 m the left charge contributes +3600 N C⁻¹, while the right contributes −9.0/(0.20²) = −225 N C⁻¹.
- 2
Net field is +3375 N C⁻¹. F = qE = (−2.0×10⁻⁹)(3375) = −6.75×10⁻⁶ N; a negative charge feels force opposite the field.
- 3
Lines point away from positive charges, never cross and are denser where field is stronger in a consistent drawing. They represent a continuous vector field, not discrete channels carrying test charges.
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.
- Between two positive charges their fields oppose. Equality requires 4/x² = 1/(0.30−x)², giving x/(0.30−x) = 2.
- Solve x = 0.20 m, with distances 0.20 and 0.10 m. Field directions, not just magnitudes, must be checked.
- Potential adds as a scalar: V = 9.0×10⁹[(4.0×10⁻⁹)/0.20+(1.0×10⁻⁹)/0.10] = 270 V. Zero field does not require zero potential.
- At x = 0.10 m the left charge contributes +3600 N C⁻¹, while the right contributes −9.0/(0.20²) = −225 N C⁻¹.
- Net field is +3375 N C⁻¹. F = qE = (−2.0×10⁻⁹)(3375) = −6.75×10⁻⁶ N; a negative charge feels force opposite the field.
- Lines point away from positive charges, never cross and are denser where field is stronger in a consistent drawing. They represent a continuous vector field, not discrete channels carrying test charges.
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