AP Physics · course AP C:E&M
AP Physics C: Electricity and Magnetism
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
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Finding the location where a field vanishes
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]
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.
Show a hint
Begin with part (a): Locate the zero-resultant-field point between the charges and find the electric potential there.
Reveal the worked solution
- 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.
Answer (a) Zero field at x = 0.20 m; potential there = +270 V. (b) Resultant field = 3375 N C⁻¹ east; force = 6.75×10⁻⁶ N west.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
A discharging capacitor transfers finite energy to a resistor
A 100 μF capacitor initially at 10.0 V discharges through a 20.0 kΩ resistor. The supplied voltages follow V(t) = V0 exp(−t/RC). There is no power supply during discharge. Use ln2 = 0.693 and neglect leakage and meter loading. Supplied data (Time / s; Voltage / V): (0, 10), (1.386, 5), (2.772, 2.5), (4.158, 1.25). (a) Find the time constant, initial current, initial stored energy and time for voltage to reach half its initial value. [3 marks] (b) Find energy remaining and energy dissipated when voltage halves. Derive the energy decay law and explain why current also halves at that time. [3 marks]
Capacitor discharge voltage
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Voltage / V
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Voltage / V |
|---|---|
| 0 | 10 |
| 1.386 | 5 |
| 2.772 | 2.5 |
| 4.158 | 1.25 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the time constant, initial current, initial stored energy and time for voltage to reach half its initial value.
Reveal the worked solution
- The time constant is RC = (2.0×10⁴)(1.0×10⁻⁴) = 2.0 s; initial current magnitude is V0/R = 10/20000 = 0.50 mA.
- Initial energy is ½CV0² = ½(1.0×10⁻⁴)(100) = 0.0050 J.
- Setting V/V0 = 1/2 gives t = RC ln2 = 1.386 s, consistent with the supplied data. The time constant is not the voltage half-time.
- Since U = ½CV² and C is fixed, halving voltage quarters stored energy, leaving 5.0/4 = 1.25 mJ.
- The resistor receives the missing 3.75 mJ as thermal energy. Substitution of V(t) gives U(t) = U0 exp(−2t/RC), so energy decays with a time constant RC/2.
- Ohm's law gives I(t) = V(t)/R, so current halves when voltage halves. Integrating I²R over the full discharge returns the finite initial capacitor energy.
Answer (a) RC = 2.0 s; Iinitial = 0.50 mA; Uinitial = 5.0 mJ; voltage half-time = 1.386 s. (b) Remaining energy = 1.25 mJ; dissipated = 3.75 mJ; U = U0 exp(−2t/RC).
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.