AP Physics · topic 10
Conductors and Capacitors
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.
A conductor's zero internal field and nonzero potential
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]
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.
Show a hint
Begin with part (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.
Reveal the worked solution
- 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.
Answer (a) Inside E = 0 and V = 180 V; just outside E = 900 N C⁻¹ radially outward. (b) External work = 1.8×10⁻¹⁰ J; capacitance≈22.2 pF.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
An isolated capacitor is not a constant-voltage capacitor
A parallel-plate capacitor of capacitance 4.0 μF is charged to 12 V and then completely disconnected from its supply. Its plate separation is subsequently doubled with no leakage. The initial charge–voltage relation is supplied. Ignore edge effects and use C = ε₀A/d. Supplied data (Potential difference / V; Stored charge / μC): (0, 0), (4, 16), (8, 32), (12, 48). (a) Find charge and stored energy before the change, then capacitance, voltage and energy after doubling the gap. [3 marks] (b) Repeat the predictions if the 12 V supply instead remains connected and explain where energy goes. [3 marks]
Initial capacitor charging relation
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Stored charge / μC
Horizontal axis (x): Potential difference / V
Read the exact data as a table
| Potential difference / V | Stored charge / μC |
|---|---|
| 0 | 0 |
| 4 | 16 |
| 8 | 32 |
| 12 | 48 |
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 and stored energy before the change, then capacitance, voltage and energy after doubling the gap.
Reveal the worked solution
- Initial charge is CV = 4.0 μF×12 V = 48 μC. Initial energy is ½CV² = 288 μJ.
- Doubling d halves capacitance to 2.0 μF. Disconnection fixes charge, so V = Q/C doubles to 24 V.
- At fixed Q, U = Q²/(2C), so halving C doubles energy to 576 μJ. External mechanical work against attraction supplies the 288 μJ increase.
- The supply fixes potential difference rather than charge. With half the capacitance, Q = CV halves to 24 μC.
- Stored field energy ½CV² also halves to 144 μJ, but this does not mean separating attracting plates releases mechanical work.
- Quasistatic separation still needs positive mechanical work. Charge returns energy to the supply; the field energy, mechanical work and supply energy must all be included in the energy balance.
Answer (a) Initially Q = 48 μC and U = 288 μJ; afterward C = 2.0 μF, V = 24 V and U = 576 μJ. (b) Connected case: C = 2.0 μF, V = 12 V, Q = 24 μC and U = 144 μJ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.