Cambridge International AS & A Level · 18 · structured
Spherical charge, Coulomb force and potential gradient
Spherical charge, Coulomb force and potential gradient · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 9
- Topics
- 5
- Answer
- Complete
An isolated spherical conducting source of radius 0.050 m carries Q=−3.00 nC. A small +2.00 nC test charge is at distance r=0.200 m from its centre. Use 1/(4πε₀)=8.99×10⁹ N m² C⁻². Assume the test charge does not distort the source distribution.
- (a)
Explain the point-charge replacement valid here and calculate the force magnitude and direction on the test charge using Coulomb's law.
3 marks - (b)
Calculate potential V at the test point relative to infinity and hence the pair's potential energy. Explain why both signs are negative.
3 marks - (c)
In a separate one-dimensional uniform field, potential falls linearly from +12.0 V at x=0.010 m to +4.0 V at x=0.030 m. Sketch V against x and obtain E_x using the negative gradient.
3 marks
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(a)
Explain the point-charge replacement valid here and calculate the force magnitude and direction on the test charge using Coulomb's law.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
The test point is outside the spherical conductor; its exterior field is equivalent to that of its charge at the centre under the stated symmetry assumption.
- 2
F=k|Qq|/r²=8.99×10⁹×3.00×10⁻⁹×2.00×10⁻⁹/0.200²=1.35×10⁻⁶ N.
- 3
Opposite charges attract, so force is radially towards the source.
(b)
Calculate potential V at the test point relative to infinity and hence the pair's potential energy. Explain why both signs are negative.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
V=kQ/r=8.99×10⁹(−3.00×10⁻⁹)/0.200=−134.85 V.
- 2
E_p=qV=2.00×10⁻⁹×(−134.85)=−2.697×10⁻⁷ J.
- 3
External work to bring the positive charge slowly from infinity is negative because the attractive field supplies energy.
(c)
In a separate one-dimensional uniform field, potential falls linearly from +12.0 V at x=0.010 m to +4.0 V at x=0.030 m. Sketch V against x and obtain E_x using the negative gradient.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
Draw the descending straight line through the two labelled coordinates.
- 2
dV/dx=(4.0−12.0)/(0.030−0.010)=−400 V m⁻¹.
- 3
E_x=−dV/dx=+400 V m⁻¹: the field points towards decreasing potential.
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.
- (a) The test point is outside the spherical conductor; its exterior field is equivalent to that of its charge at the centre under the stated symmetry assumption.
- (a) F=k|Qq|/r²=8.99×10⁹×3.00×10⁻⁹×2.00×10⁻⁹/0.200²=1.35×10⁻⁶ N.
- (a) Opposite charges attract, so force is radially towards the source.
- (b) V=kQ/r=8.99×10⁹(−3.00×10⁻⁹)/0.200=−134.85 V.
- (b) E_p=qV=2.00×10⁻⁹×(−134.85)=−2.697×10⁻⁷ J.
- (b) External work to bring the positive charge slowly from infinity is negative because the attractive field supplies energy.
- (c) Draw the descending straight line through the two labelled coordinates.
- (c) dV/dx=(4.0−12.0)/(0.030−0.010)=−400 V m⁻¹.
- (c) E_x=−dV/dx=+400 V m⁻¹: the field points towards decreasing potential.
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