understand that an electric field is an example of a field of force and define electric field as force per unit positive charge

Electric Fields and Field Lines – A‑Level Physics 9702 (Section 18.1)

Learning Objective

Understand that an electric field is a type of field of force and be able to define it as the force per unit positive test charge.

1. What Is a Field?

• A field is a region of space in which a force can be experienced by a test object placed within it.
• Common examples:

  • Gravitational field – force on a mass.
  • Electric field – force on an electric charge.
  • Magnetic field – force on a moving charge or magnetic dipole.

2. Definition of the Electric Field

The electric field E at a point is defined as the force F experienced by a small positive test charge q₀ placed at that point, divided by the magnitude of the test charge:

\[

\mathbf{E}= \frac{\mathbf{F}}{q_{0}}

\]

• Because the test charge is taken to be positive, the direction of E is the direction of the force on a positive charge.
• The field exists even when no test charge is present; the test charge is only a tool for measurement.

3. Key Properties of Electric Fields

4. Electric Field of a Point Charge (Coulomb’s Law)

For a point charge Q, the magnitude of the electric field at a distance r is

\[

E = \frac{1}{4\pi\varepsilon_{0}}\frac{|Q|}{r^{2}} \qquad \left(k = 8.99\times10^{9}\,\text{N m}^{2}\text{C}^{-2}\right)

\]

• Direction: radially outward if Q > 0, radially inward if Q < 0.

5. Drawing and Interpreting Electric Field Lines

Field lines are a visual aid that represent the direction and relative strength of E. They are not physical objects.

Checklist for Sketching Field Lines

1. Start on positive charges and end on negative charges (or extend to infinity if only one sign is present).
2. The tangent to a line at any point gives the direction of E there.
3. Line density (lines per unit area) is proportional to the magnitude of the field.
4. Lines never cross – crossing would imply two different directions for E at the same point.
5. Use arrows on the lines to indicate the direction of the field (away from +, toward –).

Typical Configurations

1. Single positive charge +Q – radial lines radiate outward uniformly.
2. Single negative charge –Q – radial lines converge inward uniformly.
3. Electric dipole (equal opposite charges) – lines emerge from +Q, curve, and terminate on –Q; density is greatest between the charges.
4. Two like charges (+Q, +Q) – lines emerge from each charge and repel each other, creating a region of low line density between them.

Sketching Exercise

Draw the field‑line diagrams for the four configurations listed above. For each diagram:

• Indicate the direction of the field with arrows.
• Show that line density is greatest close to the charges and diminishes with distance.
• Verify that no lines cross.

6. Worked Examples

Example 1 – Superposition for Two Opposite Charges

Problem: Two point charges, +5 µC and –5 µC, are 0.10 m apart. Find the magnitude and direction of the electric field at the midpoint.

1. Distance from each charge to the midpoint: \(r = 0.05\;\text{m}\).
2. Magnitude of the field due to each charge:

   \[

   E{+}=E{-}= \frac{k\,|Q|}{r^{2}}=

   \frac{8.99\times10^{9}\times5\times10^{-6}}{(0.05)^{2}}

   =1.80\times10^{5}\;\text{N C}^{-1}

   \]

3. Direction:

   • Field from the +5 µC points away from the charge → to the right.
   • Field from the –5 µC points toward the charge → also to the right.

4. Superpose (add) the collinear vectors:

   \[

   E{\text{net}} = E{+}+E{-}=2E{+}=3.6\times10^{5}\;\text{N C}^{-1}

   \]

   The net field points horizontally to the right.

Example 2 – Vector Superposition for Three Non‑Collinear Charges

Problem: Three point charges are placed as follows:

• +Q at the origin (0, 0) m
• –Q at (0, 0.20) m
• +Q at (0.20, 0) m

All have magnitude \(Q = 2.0\;\mu\text{C}\). Find the electric field at point P (0.10 m, 0.10 m).

1. Calculate the vector displacement from each charge to P and its magnitude \(r_i\).
   

   \[

   \begin{aligned}

   \mathbf{r}{1}&=(0.10,0.10)\;\text{m},\; r{1}= \sqrt{0.10^{2}+0.10^{2}}=0.141\;\text{m}\\

   \mathbf{r}{2}&=(0.10,-0.10)\;\text{m},\; r{2}=0.141\;\text{m}\\

   \mathbf{r}{3}&=(-0.10,0.10)\;\text{m},\; r{3}=0.141\;\text{m}

   \end{aligned}

   \]

2. Unit vectors \(\hat{\mathbf{r}}{i}=\mathbf{r}{i}/r_{i}\).
   

   \[

   \hat{\mathbf{r}}_{1}= (0.707,0.707),\;

   \hat{\mathbf{r}}_{2}= (0.707,-0.707),\;

   \hat{\mathbf{r}}_{3}= (-0.707,0.707)

   \]

3. Magnitude of the field from each charge (using Coulomb’s law):

   \[

   Ei = \frac{k\,Q}{ri^{2}}=

   \frac{8.99\times10^{9}\times2.0\times10^{-6}}{(0.141)^{2}}

   =9.0\times10^{5}\;\text{N C}^{-1}

   \]

4. Direction of each field:

   • From +Q at the origin → away from the charge → along \(\hat{\mathbf{r}}{1}\).
     \(\mathbf{E}{1}=Ei\,\hat{\mathbf{r}}{1}\)
   • From –Q at (0, 0.20) m → toward the charge → opposite to \(\hat{\mathbf{r}}{2}\).
     \(\mathbf{E}{2}= -Ei\,\hat{\mathbf{r}}{2}\)
   • From +Q at (0.20, 0) m → away from the charge → opposite to \(\hat{\mathbf{r}}{3}\).
     \(\mathbf{E}{3}= -Ei\,\hat{\mathbf{r}}{3}\)

5. Component‑wise addition:

   \[

   \begin{aligned}

   \mathbf{E}{\text{net}} &= \mathbf{E}{1}+\mathbf{E}{2}+\mathbf{E}{3}\\

   &=E_i\big[(0.707,0.707) - (0.707,-0.707) - (-0.707,0.707)\big]\\

   &=E_i\,(0.707,0.707 -0.707,0.707 +0.707,-0.707)\\

   &=E_i\,(0.707,0.707)\\

   &= (9.0\times10^{5})(0.707,0.707)\\

   &= (6.36\times10^{5},\;6.36\times10^{5})\;\text{N C}^{-1}

   \end{aligned}

   \]

6. Result: The electric field at P has magnitude

   \[

   |\mathbf{E}_{\text{net}}| = \sqrt{(6.36\times10^{5})^{2}+(6.36\times10^{5})^{2}}

   =9.0\times10^{5}\;\text{N C}^{-1}

   \]

   and points at 45° above the +x‑axis (i.e. along the line \(y=x\)).

7. Summary – Key Points to Remember

• The electric field is a vector field defined as \(\mathbf{E}= \mathbf{F}/q_{0}\) for a positive test charge.
• Units: N C⁻¹ or V m⁻¹.
• It is produced by static electric charges (Coulomb’s law).
• Superposition: the net field is the vector sum of the individual fields.
• Field lines are a convenient visual representation; they obey the checklist above and never intersect.
• Field lines are denser where the field is stronger and are perpendicular to equipotential surfaces.

8. Practice Questions

1. Sketch the field‑line diagrams for a single +Q, a single –Q, an electric dipole, and two like charges. Label the direction of the field and comment on line density.
2. A test charge \(q_{0}=2\,\text{nC}\) experiences a force of \(3\times10^{-6}\,\text{N}\) to the right. Calculate the electric field at that point.
3. Explain, using the definition of the electric field, why field lines never intersect.
4. Two point charges, +4 µC at (0, 0) and –2 µC at (0, 0.30) m, are present. Determine the magnitude and direction of the electric field at the point (0.20 m, 0). (Hint: resolve each contribution into x‑ and y‑components and then add.)