AP · AP C:E&M · extended response
A conductor's zero internal field and nonzero potential
AP Physics C: Electricity and Magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 3
- Answer
- Complete
Exterior shell potential
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Electric potential / V
Horizontal axis (x): Radius / m
Read the exact data as a table
| Radius / m | Electric potential / V |
|---|---|
| 0.2 | 180 |
| 0.4 | 90 |
| 0.6 | 60 |
| 0.8 | 45 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Choose your AP course · School year 2026–27 · May 2027 exam
Choose a course to check this question’s reviewed parts and required depth. Its original AP label and topic codes do not establish suitability for the current syllabus.
Choose reviewed AP practice →An isolated spherical conducting shell has radius 0.20 m and charge +4.0 nC. Use k = 9.0×10⁹ N m² C⁻² and zero potential at infinity. There is no charge inside the cavity. The supplied graph shows the measured potential at exterior radii; the conductor is in electrostatic equilibrium. Supplied data (Radius / m; Electric potential / V): (0.2, 180), (0.4, 90), (0.6, 60), (0.8, 45).
- (a)
Use Gauss's law to state the field inside and outside the shell; find potential at the centre and field just outside the surface.
3 marks - (b)
Find work required by an external agent to bring a +1.0 pC test charge slowly from infinity to the centre, and infer the shell capacitance.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Use Gauss's law to state the field inside and outside the shell; find potential at the centre and field just outside the surface.
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 spherical Gaussian surface outside encloses Q, so E(4πr²) = Q/ε₀ and E = kQ/r². Spherical symmetry is essential to taking E outside the integral.
- 2
The empty cavity and conductor have zero electrostatic field. Since E = −dV/dr, potential is constant inside and equals the surface value kQ/R = 180 V.
- 3
Just outside, E = 9.0×10⁹(4.0×10⁻⁹)/(0.20²) = 900 N C⁻¹ outward. Zero internal field does not set internal potential to zero.
(b)
Find work required by an external agent to bring a +1.0 pC test charge slowly from infinity to the centre, and infer the shell capacitance.
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
For a sufficiently small test charge that negligibly disturbs the shell model, ΔU = qΔV = (1.0×10⁻¹²)(180) = 1.8×10⁻¹⁰ J.
- 2
Moving from the surface to the centre requires no additional electrostatic work because potential is constant; the work occurs on the exterior approach.
- 3
The isolated sphere's capacitance is Q/V = 4.0×10⁻⁹/180≈2.22×10⁻¹¹ F. Capacitance is fixed by geometry and surroundings rather than the particular stored charge in this linear model.
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 spherical Gaussian surface outside encloses Q, so E(4πr²) = Q/ε₀ and E = kQ/r². Spherical symmetry is essential to taking E outside the integral.
- The empty cavity and conductor have zero electrostatic field. Since E = −dV/dr, potential is constant inside and equals the surface value kQ/R = 180 V.
- Just outside, E = 9.0×10⁹(4.0×10⁻⁹)/(0.20²) = 900 N C⁻¹ outward. Zero internal field does not set internal potential to zero.
- For a sufficiently small test charge that negligibly disturbs the shell model, ΔU = qΔV = (1.0×10⁻¹²)(180) = 1.8×10⁻¹⁰ J.
- Moving from the surface to the centre requires no additional electrostatic work because potential is constant; the work occurs on the exterior approach.
- The isolated sphere's capacitance is Q/V = 4.0×10⁻⁹/180≈2.22×10⁻¹¹ F. Capacitance is fixed by geometry and surroundings rather than the particular stored charge in this linear model.
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