recall and use F = qE for the force on a charge in an electric field

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?

• Definition (vector form) – The electric field $$\mathbf{E}$$ at a point is the force per unit positive test charge placed at that point:

  $$\displaystyle \mathbf{E}=\frac{\mathbf{F}}{q_0}$$

  where $$q_0$$ is a small positive charge (so that it does not disturb the existing field).
• Units – newtons per coulomb (N C⁻¹) or volts per metre (V m⁻¹). Both are equivalent because $$1\;\text{N C}^{-1}=1\;\text{V m}^{-1}$$.
• Direction – given by the direction of the force on a positive test charge; therefore the field is a **vector** quantity.

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

1. Convert the charge: $$Q = 5.0\times10^{-6}\;\text{C}$$.
2. Insert values:

   $$E = \frac{9.0\times10^{9}\;(5.0\times10^{-6})}{(0.040)^{2}} = \frac{4.5\times10^{4}}{1.6\times10^{-3}} \approx 2.8\times10^{4}\;\text{N C}^{-1}$$

3. 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)

• Potential of a point charge: $$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$
• Potential energy of a test charge: $$U = qV$$
• Potential is a scalar; it can be added algebraically, unlike vectors.

Example – Potential energy of an electron near a +2.0 µC charge

1. Potential at 0.10 m: $$V = \frac{9.0\times10^{9}(2.0\times10^{-6})}{0.10}=1.8\times10^{4}\;\text{V}$$
2. Potential energy: $$U = (-1.60\times10^{-19})(1.8\times10^{4}) = -2.9\times10^{-15}\;\text{J}$$
3. 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

10. Practice Questions (Cambridge‑style)

1. 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.
2. 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 ($$q = +1.60\times10^{-19}\;\text{C}$$)?
3. 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.
4. 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.
5. 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)

1. 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 $$\mathbf{E}_{\text{net}} = 0$$.
2. $$F = qE = (1.60\times10^{-19})(2.0\times10^{4}) = 3.2\times10^{-15}\;\text{N}$$ upward.
3. 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.
4. $$E = V/d = 150\;\text{V}/(3.0\times10^{-3}\;\text{m}) = 5.0\times10^{4}\;\text{N C}^{-1}$$. Force on electron: $$F = (-1.60\times10^{-19})(5.0\times10^{4}) = -8.0\times10^{-15}\;\text{N}$$ (opposite to the field direction).
5. 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: $$V = kQ/r,\; U = qV$$.