recall and use Coulomb’s law F = Q1Q2 / (4πε0 r 2) for the force between two point charges in free space

Electric Fields and Coulomb’s Law – Cambridge IGCSE / A‑Level Physics (9702 – Section 18)

1. Electric Field

• Definition: The electric field 𝐄 at a point is the force per unit positive test charge placed at that point

  \$\mathbf{E}=\frac{\mathbf{F}}{q_{\text{test}}}\qquad\bigl[\text{N C}^{-1}= \text{V m}^{-1}\bigr]\$

• Vector quantity: has magnitude and direction.

  • Direction = direction of the force that would act on a positive test charge.
  • For a negative test charge the force is opposite to the field direction.

• Field‑line conventions (qualitative tool):

  • Lines start on positive charges and end on negative charges.
  • Density of lines ∝ magnitude of 𝐄.
  • Lines never cross.

2. Field of a Point Charge

For an isolated point charge Q the field is radial and follows an inverse‑square law:

\$\$\mathbf{E}= \frac{1}{4\pi\varepsilon_{0}}\,\frac{Q}{r^{2}}\;\hat{\mathbf{r}}

\;=\;k\frac{Q}{r^{2}}\;\hat{\mathbf{r}} \qquad\bigl(k=8.99\times10^{9}\,\text{N m}^{2}\text{C}^{-2}\bigr)\$\$

• Direction: outward if Q > 0, inward if Q < 0.
• The force on a second charge q at distance r is obtained from 𝐅 = q𝐄, which is equivalent to Coulomb’s law.

3. Uniform Electric Fields

3.1 Infinite Charged Sheet

For an infinitely large sheet with surface charge density σ (C m⁻²), Gauss’s law gives a constant field on each side:

\$E{\text{sheet}}=\frac{\sigma}{2\varepsilon{0}}\quad\text{(uniform, perpendicular to the sheet)}\$

• Magnitude does not depend on distance from the sheet.
• Direction: away from the sheet if σ > 0, toward the sheet if σ < 0.

3.2 Parallel‑Plate Approximation

Two large parallel plates carrying opposite surface charge densities +σ and –σ produce a nearly uniform field between them because the fields from the two sheets add:

\$E{\text{between}}=\frac{\sigma}{\varepsilon{0}}\$

In practice we use the simpler relation

\$E=\frac{V}{d}\qquad\text{or}\qquad V=E\,d\$

where d is the plate separation and V the potential difference.

• Field lines are straight, parallel and equally spaced (ideal case).
• Edge effects: near the plate edges the lines bend; the uniform‑field approximation is valid only far from the edges (the “large‑plate” condition).

4. Coulomb’s Law – Force Between Point Charges

The magnitude of the electrostatic force between two point charges Q₁ and Q₂ separated by a distance r in free space is

\$F = \frac{1}{4\pi\varepsilon{0}}\,\frac{|Q{1}Q{2}|}{r^{2}} = k\frac{|Q{1}Q_{2}|}{r^{2}}\$

Vector form (including direction):

\$\mathbf{F}{12}=k\frac{Q{1}Q{2}}{r^{2}}\;\hat{\mathbf{r}}{12}\$

• If Q₁Q₂ > 0 the force is repulsive (directed along \$\hat{\mathbf{r}}_{12}\$).
• If Q₁Q₂ < 0 the force is attractive (directed opposite to \$\hat{\mathbf{r}}_{12}\$).
• Newton’s III law: \$\mathbf{F}{12}= -\mathbf{F}{21}\$.
• Superposition principle: for more than two charges the net force on any charge is the vector sum of the individual forces from each other charge.

Link to the Field Concept

The field produced by Q₁ at the location of Q₂ is

\$\mathbf{E}{1}=k\frac{Q{1}}{r^{2}}\;\hat{\mathbf{r}}_{12}\$

and the force on Q₂ is simply

\$\mathbf{F}{12}=Q{2}\mathbf{E}_{1}\$

This shows that Coulomb’s law is a special case of the general relation \$\mathbf{F}=q\mathbf{E}\$.

5. Electric Potential

5.1 Definition

Electric potential V at a point is the work required to bring a unit positive charge from infinity to that point, without acceleration:

\$V=\frac{W}{q}\qquad\bigl[\text{J C}^{-1}= \text{V}\bigr]\$

Potential is a scalar; the potential difference between points A and B is

\$\Delta V = V{B}-V{A}= -\int_{A}^{B}\mathbf{E}\!\cdot\!d\mathbf{s}\$

5.2 Potential of a Point Charge

\$V(r)=k\frac{Q}{r}\$

Reference is usually taken at infinity where \$V=0\$.

5.3 Uniform Field

When the field is uniform and directed along the \$x\$‑axis:

\$\Delta V = -E\,\Delta x\qquad\text{or}\qquad V = -Ex + \text{constant}\$

5.4 Equipotential Surfaces

• All points on an equipotential have the same potential.
• Equipotentials are always perpendicular to electric‑field lines.
• For a uniform field they are a set of parallel planes spaced by

  \$\Delta x = \frac{\Delta V}{E}\$

6. Worked Example – Coulomb’s Law

Problem: Two point charges, \$Q{1}=+5.0\,\mu\text{C}\$ and \$Q{2}=-3.0\,\mu\text{C}\$, are \$r=0.20\,\$m apart in vacuum. Find the magnitude and direction of the force on each charge.

1. Convert to coulombs: \$Q{1}=5.0\times10^{-6}\,\$C, \$Q{2}=-3.0\times10^{-6}\,\$C.
2. Magnitude (Coulomb’s law):

   \$\$F = k\frac{|Q{1}Q{2}|}{r^{2}}

   = 8.99\times10^{9}\frac{(5.0\times10^{-6})(3.0\times10^{-6})}{(0.20)^{2}}

   \approx 1.35\ \text{N}\$\$

3. Direction: \$Q{1}Q{2}<0\$ ⇒ the force is attractive.

   • On \$Q{1}\$ the force points toward \$Q{2}\$.
   • On \$Q{2}\$ the force points toward \$Q{1}\$.
   • Both forces have the same magnitude (Newton’s III).

7. Worked Example – Uniform Field and Potential

Problem: A parallel‑plate capacitor has plate separation \$d=4.0\,\$cm and a potential difference \$V=120\,\$V. Determine (a) the field magnitude, (b) the force on an electron placed midway, and (c) the electron’s initial acceleration.

1. Field magnitude:

   \$E = \frac{V}{d}= \frac{120}{0.040}=3.0\times10^{3}\ \text{N C}^{-1}\$

2. Force on an electron (\$q=-e=-1.60\times10^{-19}\,\$C):

   \$\$F = qE = (-1.60\times10^{-19})(3.0\times10^{3})

   = -4.8\times10^{-16}\ \text{N}\$\$

   (negative sign = opposite to the field direction).

3. Acceleration (\$m_{e}=9.11\times10^{-31}\,\$kg):

   \$\$a = \frac{|F|}{m_{e}} = \frac{4.8\times10^{-16}}{9.11\times10^{-31}}

   \approx 5.3\times10^{14}\ \text{m s}^{-2}\$\$

8. Practice Questions

1. Two identical point charges \$+2.0\,\mu\text{C}\$ separated by \$0.10\,\$m.

   
   \$F = k\frac{(2.0\times10^{-6})^{2}}{(0.10)^{2}}\approx 3.6\ \text{N}\$ (repulsive).

2. Parallel‑plate capacitor \$d=5.0\,\$cm, \$V=200\,\$V.

   
   \$E = \frac{V}{d}= \frac{200}{0.050}=4.0\times10^{3}\ \text{N C}^{-1}\$

3. Proton in a uniform field \$E=5.0\times10^{5}\ \text{N C}^{-1}\$.

   
   Force: \$F=eE= (1.60\times10^{-19})(5.0\times10^{5})=8.0\times10^{-14}\ \text{N}\$

   
   Acceleration: \$a=F/m_{p}=8.0\times10^{-14}/1.67\times10^{-27}\approx4.8\times10^{13}\ \text{m s}^{-2}\$.

4. Potential of a point charge: \$Q=+8.0\,\$nC at the origin, find \$V\$ at \$r=0.15\,\$m.

   
   \$V = k\frac{Q}{r}= \frac{8.99\times10^{9}\times8.0\times10^{-9}}{0.15}\approx4.8\times10^{2}\ \text{V}\$

5. Superposition: Three charges on a line: \$Q{1}=+1\,\mu\text{C}\$ at \$x=0\$, \$Q{2}=-2\,\mu\text{C}\$ at \$x=0.05\,\$m, \$Q{3}=+3\,\mu\text{C}\$ at \$x=0.12\,\$m. Find the net force on \$Q{2}\$.

   
   Solution sketch:

   • \$\mathbf{F}{12}=k\frac{Q{1}Q_{2}}{(0.05)^{2}}\$ (attractive, to the left).
   • \$\mathbf{F}{32}=k\frac{Q{3}Q_{2}}{(0.07)^{2}}\$ (attractive, to the right).
   • Vector sum gives \$F_{\text{net}}\approx2.1\ \text{N}\$ toward the left.

9. Summary Table

10. Key Points to Remember

• Coulomb’s law applies only to point charges (or spherically symmetric charge distributions) in free space.
• Electric field is a vector; electric potential is a scalar.
• For a uniform field \$E=V/d\$; for a point charge \$E=kQ/r^{2}\$ and \$V=kQ/r\$.
• Superposition: add forces (vectors) or fields (vectors) component‑by‑component.
• Equipotential surfaces are always perpendicular to field lines.
• Edge effects become significant when the plate dimensions are not much larger than the separation.