understand that, for a point outside a spherical conductor, the charge on the sphere may be considered to be a point charge at its centre

Electric Fields – Spherical Conductors and Uniform Fields (Cambridge 9702 – 18.1 – 18.5)

Learning Objective

Explain why a point outside a charged spherical conductor can be treated as if the whole charge were a point charge at its centre, and use this idea together with Gauss’s law, Coulomb’s law and the principle of superposition to analyse uniform electric fields.

Key Concepts

  • Electric field E – vector defined by the force on a positive test charge: F = q E.
  • Field‑line representation – direction = tangent to a line, density ∝ |E|, lines start on positive charges and end on negative charges (or at infinity). Field lines are a *convention*; they are not physical objects.
  • Electric flux ΦE – ΦE = ∮ E·dA = E A cosθ = Qenc/ε₀ (Gauss’s law). The proportionality between line density and flux links the visual picture to the quantitative law.
  • Coulomb’s law (point charges)

    \$\mathbf{F}=k\frac{Q1Q2}{r^{2}}\hat{\mathbf{r}},\qquad k=\frac{1}{4\pi\varepsilon_0}\$

    Repulsive if the signs are the same, attractive otherwise.

  • Electric field of a point charge – from F = qE and Coulomb’s law:

    \$\mathbf{E}(r)=k\frac{Q}{r^{2}}\hat{\mathbf{r}}.\$

  • Point‑charge equivalence of a spherical conductor – a charged conducting sphere produces the same external field as a point charge Q at its centre.
  • Superposition principle – the resultant field at any point is the vector sum of the fields produced by each charge distribution.
  • Uniform electric field – constant magnitude and direction throughout a region; ideally realised by two large parallel plates with E = ΔV/d.

1. Electric Fields & Field Lines (Syllabus 18.1)

  • Definition:  E = F/q (force per unit positive charge).
  • Direction: along the force on a positive test charge; opposite for a negative test charge.
  • Field‑line rules

    • Lines never cross – crossing would give two directions for E at the same point.
    • Number of lines per unit area is proportional to the magnitude of E (density ∝ |E|).
    • Lines start on positive charges and end on negative charges (or at infinity).
    • They are a visual convention, not physical entities.

  • Quantitative link to flux: For a small surface element dA, the flux is Φ = E·dA = E A cosθ. Summing over a closed surface gives Gauss’s law.

2. Electric Force Between Point Charges (Syllabus 18.3)

Coulomb’s law in vector form:

\$\mathbf{F}=k\frac{Q1Q2}{r^{2}}\hat{\mathbf{r}}\$

  • k = 1/(4π ε₀) = 9.0 × 10⁹ N m² C⁻².
  • Direction is along the line joining the charges; repulsive if Q₁Q₂ > 0, attractive if Q₁Q₂ < 0.

3. Electric Field of a Point Charge (Syllabus 18.4)

From F = qE and Coulomb’s law:

\$\mathbf{E}(r)=\frac{\mathbf{F}}{q}=k\frac{Q}{r^{2}}\hat{\mathbf{r}}\$

  • Radial, pointing away from a positive charge (toward a negative charge).
  • Magnitude falls as 1/r².

4. Spherical Conductor – Point‑Charge Equivalence (Syllabus 18.2 & 18.5)

Consider a conducting sphere of radius R carrying total charge Q.

  1. Gaussian surface: a sphere centred on the conductor with radius r > R.
  2. Symmetry: the field on the Gaussian surface is radial and has the same magnitude E(r) everywhere.
  3. Gauss’s law:

    \$\oint\mathbf{E}\!\cdot\!d\mathbf{A}=E(r)\,4\pi r^{2}= \frac{Q}{\varepsilon_{0}}.\$

  4. Resulting field:

    \$E(r)=k\frac{Q}{r^{2}},\qquad \mathbf{E}(r)=k\frac{Q}{r^{2}}\hat{\mathbf{r}}.\$

    This is identical to the field of a point charge Q at the centre.

Consequences:

  • Outside the sphere (r > R) – behaves exactly like a point charge at the centre.
  • Inside the conductor (r < R) – electrostatic equilibrium forces the net electric field to zero; excess charge resides only on the outer surface.
  • The surface of the conductor is an equipotential, so the potential is the same at every point on the sphere.

5. Uniform Electric Fields (Syllabus 18.2)

  • Definition: E has the same magnitude and direction at every point in the region of interest.
  • Ideal realisation: two large parallel conducting plates, area A ≫ d (plate dimensions far greater than the separation). The field between them is

    \$\mathbf{E}= \frac{ΔV}{d}\,\hat{\mathbf{n}}\$

    where ΔV is the potential difference and \(\hat{\mathbf{n}}\) is a unit vector normal to the plates.

  • Edge effects: Near the plate edges the field lines diverge, reducing uniformity. In the central region (away from edges) the field can be treated as uniform for A‑Level calculations.
  • Validity condition: uniform‑field approximation is reliable when the plate dimensions are at least an order of magnitude larger than the separation d.

6. Superposition and Construction of Approximately Uniform Fields

The electric field obeys the principle of superposition:

\$\mathbf{E}{\text{total}}=\sumi \mathbf{E}_i.\$

Two useful syllabus‑aligned examples:

6.1 Parallel‑plate capacitor

  • Each plate can be replaced by a sheet of charge with surface charge density σ = Q/A.
  • The field due to a single infinite sheet is E = σ/(2ε₀) (directed away from a positive sheet).
  • Superposing the fields of the two oppositely charged sheets gives a uniform field of magnitude E = σ/ε₀ = ΔV/d between the plates, and zero outside.

6.2 Two widely separated spherical conductors

  • Each sphere can be replaced by a point charge at its centre (Section 4) provided the observation point is far from both spheres (r ≫ R).
  • If the charges are equal and opposite and the line joining the centres is perpendicular to the region of interest, the vector sum of the two radial fields can be nearly constant over a small central volume – an approximation sometimes used for “parallel‑plate‑like” fields generated by large spheres.
  • Limitations: the approximation breaks down when the point is comparable to the sphere radii or when the separation of the spheres is not much larger than their radii.

7. Summary Table

RegionElectric Field EReasoning
Inside a conducting sphere (r < R)0Electrostatic equilibrium → charges reside on outer surface; conductor is equipotential.
On the surface (r = R)\(\displaystyle k\frac{Q}{R^{2}}\)Surface charge produces a radial field just outside the surface.
Outside the sphere (r > R)\(\displaystyle k\frac{Q}{r^{2}}\)Gauss’s law → identical to a point charge at the centre.
Between large parallel plates\(\displaystyle \frac{ΔV}{d}\) (constant)Uniform‑field approximation; valid when plate dimensions ≫ d.

8. Common Misconceptions

  • “Charge spreads throughout the volume of a conductor.” – In electrostatic equilibrium excess charge resides only on the outer surface.
  • “The field inside a charged sphere is the same as outside.” – Inside a *conducting* sphere the field is zero; only a uniformly charged *non‑conducting* sphere has a non‑zero internal field.
  • “The point‑charge model works only for very small spheres.” – It works for any spherical conductor as long as the observation point lies outside the sphere.
  • “Field lines are physical objects.” – They are a useful convention; the actual field exists everywhere, even where no lines are drawn.
  • “Uniform fields can be produced by any arrangement of charges.” – Only configurations where the superposed field has the same magnitude and direction over the region of interest (e.g., large parallel plates) give a truly uniform field.

9. Worked Example – Field Outside a Charged Conducting Sphere

Problem: A conducting sphere of radius R = 0.10 m carries a charge Q = +5.0 µC. Find the magnitude of the electric field 5.0 cm above the surface along the radial line.

Solution:

  1. Distance from the centre:  \(r = R + 0.05\ \text{m} = 0.15\ \text{m}\).
  2. Use the point‑charge expression (Section 4):

    \$\$E = k\frac{Q}{r^{2}}

    = \frac{9.0\times10^{9}\ \text{N·m}^{2}\!\!/\!\text{C}^{2}\times5.0\times10^{-6}\ \text{C}}{(0.15\ \text{m})^{2}}.\$\$

  3. Calculate:

    \$\$E = \frac{9.0\times10^{9}\times5.0\times10^{-6}}{0.0225}

    \approx 2.0\times10^{6}\ \text{N·C}^{-1}.\$\$

  4. The field points radially outward because the sphere is positively charged.

10. Suggested Diagrams

  • Radial field lines around a charged spherical conductor with a larger Gaussian sphere drawn around it.
  • Field‑line conventions (arrow direction, density, start/end on charges, and note that they are a convention).
  • Two large parallel plates showing straight, equally spaced field lines between them and curved lines near the edges.
  • Vector addition of the fields from two opposite point‑charge equivalents illustrating how a uniform region can arise.
  • Equipotential surfaces for a conducting sphere (concentric spheres) and for parallel plates (parallel planes).

11. Key Take‑away

For any external point, a charged spherical conductor behaves exactly like a point charge placed at its centre. This follows directly from Gauss’s law and the spherical symmetry of the conductor. By treating conductors as point charges and applying the superposition principle, we can construct and analyse uniform electric fields—most notably those produced by large parallel plates—while recognising the limits of the idealisations (edge effects, distance requirements, and equipotential nature of conductors).