AP Physics · topic 8
Electric Charges, Fields, and Gauss's Law
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.
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.