Electric Fields and the Force on a Charge – Cambridge IGCSE/A‑Level Physics (9702)
Learning Objective
Recall and use the relationship F = q E to determine the force acting on a charge placed in an electric field, and connect the field to electric potential, field‑line representation and superposition.
1. What is an Electric Field?
Experimental context
In the laboratory a tiny charged probe (often a metal sphere) is moved through a region of space. The direction of the force on the probe (measured with a torsion balance or a galvanometer) gives the direction of \$\mathbf{E}\$; the magnitude of the force divided by the known charge of the probe gives the field strength.
2. Representing Electric Fields with Field Lines
- Field lines are a visual tool; they are not physical objects.
- Rules (Cambridge syllabus 18.1):
- Lines start on positive charges and end on negative charges (or on infinity).
- The tangent to a line at any point gives the direction of \$\mathbf{E}\$ there.
- Density of lines (lines per unit area) is proportional to the magnitude of the field.
- In a uniform field the lines are parallel and equally spaced.
- Uniform field example – Two large parallel plates, the left plate +V, the right plate –V. The field is constant, directed from the positive to the negative plate, and its magnitude is \$E=V/d\$.
3. Electric Field of a Single Point Charge
For a point charge \$Q\$ the field at a distance \$r\$ is radial and given by Coulomb’s law:
\$\displaystyle \mathbf{E}= \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}\;\hat{r}\$
- \$\hat{r}\$ is a unit vector pointing away from a positive charge (toward a negative charge).
- Magnitude: \$E = k\frac{|Q|}{r^{2}}\$ with \$k = 9.0\times10^{9}\;\text{N m}^{2}\text{C}^{-2}\$.
Worked Example – Field of a +5.0 µC charge at 4.0 cm
- Convert the charge: \$Q = 5.0\times10^{-6}\;\text{C}\$.
- Insert values:
\$\$E = \frac{9.0\times10^{9}\;(5.0\times10^{-6})}{(0.040)^{2}}
= \frac{4.5\times10^{4}}{1.6\times10^{-3}}
\approx 2.8\times10^{4}\;\text{N C}^{-1}\$\$
- Direction: radially outward because the source charge is positive.
4. Superposition of Electric Fields (Syllabus 18.2)
When more than one source charge is present, the total field at a point is the vector sum of the individual fields:
\$\displaystyle \mathbf{E}{\text{net}} = \sum{i}\mathbf{E}_{i}\$
- Calculate each \$\mathbf{E}_{i}\$ using the point‑charge formula (including direction).
- Add the vectors tip‑to‑tail (or resolve into components) to obtain \$\mathbf{E}_{\text{net}}\$.
Illustration – Two equal opposite charges
Charges \$+Q\$ and \$-Q\$ are 0.10 m apart. At the midpoint the fields from each have the same magnitude but opposite direction, so \$\mathbf{E}_{\text{net}}=0\$. This is the classic dipole centre.
5. From Field to Force – F = qE
Once the electric field at a location is known, the force on any charge \$q\$ placed there is
\$\displaystyle \mathbf{F}=q\,\mathbf{E}\$
- If \$q>0\$, the force is in the same direction as the field.
- If \$q<0\$, the force is opposite to the field direction.
- Magnitude: \$F=|q|E\$.
Worked Example – Uniform field
Uniform field: \$E = 5.0\times10^{3}\;\text{N C}^{-1}\$ (rightward). Charge: \$q = -2.0\;\mu\text{C}\$.
\$F = qE = (-2.0\times10^{-6})(5.0\times10^{3}) = -1.0\times10^{-2}\;\text{N}\$
Negative sign ⇒ 0.01 N to the left.
6. Relation Between Electric Field and Electric Potential (Syllabus 18.3)
The field is the spatial rate of change of electric potential:
\$\displaystyle \mathbf{E}= -\frac{\Delta V}{\Delta d}\;\hat{n}\$
- \$\Delta V\$ is the potential difference between two points separated by \$\Delta d\$ measured along the normal \$\hat{n}\$ to an equipotential surface.
- The minus sign shows that the field points from higher to lower potential.
Uniform‑plate calculation
Plates 2.0 mm apart, \$V=500\;\text{V}\$:
\$E = \frac{V}{d}= \frac{500}{2.0\times10^{-3}} = 2.5\times10^{5}\;\text{N C}^{-1}\$
7. Electric Potential and Potential Energy (Syllabus 18.4)
Example – Potential energy of an electron near a +2.0 µC charge
- Potential at 0.10 m:
\$V = \frac{9.0\times10^{9}(2.0\times10^{-6})}{0.10}=1.8\times10^{4}\;\text{V}\$
- Potential energy:
\$U = (-1.60\times10^{-19})(1.8\times10^{4}) = -2.9\times10^{-15}\;\text{J}\$
- Negative sign indicates the electron’s energy is lower than at infinity.
8. Equipotential Lines (Syllabus 18.5)
- Equipotential lines (or surfaces) are loci of points with the same electric potential.
- They are always perpendicular to electric field lines.
- Moving a charge along an equipotential requires no work because \$\Delta V = 0 \;\Rightarrow\; \Delta U = q\Delta V = 0\$.
Sketch – Single positive point charge
Equipotentials are concentric circles (in 2‑D) centred on the charge; the closer the circle, the higher the potential.
9. Summary Table – Field Direction & Force for Common Configurations
| Configuration | Field‑line direction | Force on +q | Force on –q |
|---|
| (All forces are given by \$\mathbf{F}=q\mathbf{E}\$) |
|---|
| Single positive point charge | Radially outward | Away from the source | Toward the source |
| Single negative point charge | Radially inward | Toward the source | Away from the source |
| Uniform field (parallel plates) | From + plate to – plate | Same as field direction | Opposite to field direction |
| Electric dipole (equal opposite charges) | From + to –, curving around the pair | Depends on position; zero at the centre | Opposite to the +‑charge case |
10. Practice Questions (Cambridge‑style)
- Mid‑point field of two equal opposite charges – Two point charges, \$+3\;\mu\text{C}\$ and \$-3\;\mu\text{C}\$, are 0.10 m apart. Determine the magnitude and direction of the electric field at the midpoint.
- Force on a proton in a uniform field – A uniform field of \$2.0\times10^{4}\;\text{N C}^{-1}\$ points upward. What force does it exert on a proton (\$q = +1.60\times10^{-19}\;\text{C}\$)?
- Net field at the centre of an equilateral triangle – Three charges are placed at the vertices of an equilateral triangle of side \$0.05\;\text{m}\$: \$+2\;\mu\text{C}\$, \$+2\;\mu\text{C}\$, and \$-4\;\mu\text{C}\$. Determine the net electric field at the centre (magnitude and direction) and sketch the vector addition.
- Field and force between parallel plates – Plates are 3.0 mm apart with a potential difference of 150 V. Find the uniform field magnitude and the force on an electron placed midway.
- Work on an equipotential – A charge \$q = +5.0\times10^{-9}\;\text{C}\$ moves along an equipotential surface where the potential is 200 V. How much work is done on the charge? Explain why.
Answers (for self‑checking)
- Each charge gives \$E = kQ/r^{2}\$ with \$r = 0.05\;\text{m}\$. The two fields are equal in magnitude and opposite in direction, so \$\mathbf{E}_{\text{net}} = 0\$.
- \$F = qE = (1.60\times10^{-19})(2.0\times10^{4}) = 3.2\times10^{-15}\;\text{N}\$ upward.
- Distance from each vertex to the centre is \$d = \frac{\sqrt{3}}{3}\,0.05 \approx 0.029\;\text{m}\$. Compute the three vectors, add them tip‑to‑tail; the result is approximately \$E \approx 1.1\times10^{5}\;\text{N C}^{-1}\$ directed toward the –4 µC charge.
- \$E = V/d = 150\;\text{V}/(3.0\times10^{-3}\;\text{m}) = 5.0\times10^{4}\;\text{N C}^{-1}\$.
Force on electron: \$F = (-1.60\times10^{-19})(5.0\times10^{4}) = -8.0\times10^{-15}\;\text{N}\$ (opposite to the field direction).
- Work = \$W = q\Delta V = q(0) = 0\;\text{J}\$ because the potential does not change on an equipotential surface.
Quick Revision Checklist
- Electric field is a vector: \$\mathbf{E}= \mathbf{F}/q_0\$.
- Point‑charge field: \$\mathbf{E}=kQ/r^{2}\;\hat{r}\$.
- Uniform field: \$E = V/d\$, parallel lines.
- Superposition: add all individual \$\mathbf{E}{i}\$ vectorially before using \$\mathbf{F}=q\mathbf{E}{\text{net}}\$.
- Force direction follows the sign of \$q\$.
- Equipotential lines ⟂ field lines; moving along them requires no work.
- Potential and energy: \$V = kQ/r,\; U = qV\$.