recall and use E = Q / (4πε0 r 2) for the electric field strength due to a point charge in free space

Learning Objective (Cambridge 9702 AO1‑AO3)

Students will be able to:

• Recall the expression for the electric field of an isolated point charge in free space and its vector form.
• Use the formula to calculate the magnitude and direction of the field, to relate the field to electric force, field lines, uniform fields and electric potential, and to apply the principle of superposition for two or more point charges.
• Plan a simple experiment (e.g. using a test charge and an electroscope) to map the field of a point charge, satisfying AO3 requirements.

Prerequisite Checklist (AS‑level concepts)

Before tackling this topic, students should be comfortable with:

• Vectors and unit‑vector notation.
• Quantities, units and dimensional analysis (SI system).
• Kinematics and dynamics (forces, Newton’s II law).
• Work, energy and the relationship F·d = ΔU.
• Coulomb’s law for the force between two point charges.
• Basic concepts of electric charge, conservation of charge and the notion of a test charge.

Conceptual Overview – Electric Fields & Field Lines (Syllabus 18.1)

• Electric field \(\vec E\): a vector defined as the force \(\vec F\) experienced by a positive test charge \(q\) divided by the magnitude of that charge.
  

  \[\vec E = \frac{\vec F}{q}\]

• Direction of \(\vec E\) is the direction of the force on a positive test charge; it points away from positive charges and toward negative charges.
• Field lines are a visual representation:

  • Lines start on positive charges and end on negative charges (or extend to infinity).
  • Line density is proportional to field strength – closer lines mean a stronger field.
  • The tangent to a field line at any point gives the direction of \(\vec E\).

• For a single point charge the field is radially symmetric, leading to the inverse‑square law.

Electric Field of a Point Charge (Syllabus 18.4)

Key Quantities

• Point charge \(Q\): idealised charge treated as if all its charge were concentrated at a single point.
• Radial distance \(r\): straight‑line distance from the charge to the point where the field is evaluated (SI unit: m).
• Permittivity of free space \(\varepsilon0\): \(\displaystyle \varepsilon0 = 8.854\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}\).
• Unit vector \(\hat r\): points outward from the charge (for \(Q>0\)) or inward (for \(Q<0\)).

Derivation (from Coulomb’s law)

Coulomb’s law for the force between two point charges \(Q\) and \(q\) separated by \(r\) is

\[

\vec F = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{r^{2}}\;\hat r

\]

Dividing by the test charge \(q\) gives the field produced by \(Q\):

\[

\boxed{\displaystyle \vec E = \frac{Q}{4\pi\varepsilon_0\,r^{2}}\;\hat r}

\]

This is the vector form required by the syllabus; the scalar magnitude is obtained by taking the absolute value of \(Q\) and ignoring \(\hat r\).

Using the Formula – Step‑by‑Step

1. Identify the magnitude and sign of the point charge \(Q\) (C).
2. Measure the distance \(r\) from the charge to the point of interest (m). Square the distance!
3. Insert the values into the scalar form

   \[E = \frac{|Q|}{4\pi\varepsilon_0 r^{2}}\]

   (use the sign of \(Q\) later to set the direction).

4. Calculate \(E\); the result is in N C\(^{-1}\) (equivalent to V m\(^{-1}\)).
5. Assign direction using \(\hat r\):

   • \(Q>0\): \(\vec E\) points radially outward.
   • \(Q<0\): \(\vec E\) points radially inward.

Worked Example 1 – Positive Charge

Problem: A point charge of \(+5.0\;\mu\text{C}\) is at the origin. Find the magnitude of the electric field at a point \(0.20\;\text{m}\) from the charge.

1. Convert the charge: \(Q = 5.0\times10^{-6}\ \text{C}\).
2. Distance: \(r = 0.20\ \text{m}\).
3. Insert into the formula:

   \[

   E = \frac{5.0\times10^{-6}}{4\pi(8.854\times10^{-12})(0.20)^{2}}

   \]

4. Calculate:

   \[

   E \approx 2.24\times10^{5}\ \text{N C}^{-1}

   \]

5. Direction: radially outward from the origin (because \(Q>0\)).

Worked Example 2 – Negative Charge (Vector Form)

Problem: A point charge of \(-2.0\;\text{nC}\) is placed at the origin. Find the electric field vector at the point \((0,0,0.10\ \text{m})\).

1. Convert the charge: \(Q = -2.0\times10^{-9}\ \text{C}\).
2. Distance: \(r = 0.10\ \text{m}\).
3. Magnitude:

   \[

   E = \frac{2.0\times10^{-9}}{4\pi(8.854\times10^{-12})(0.10)^{2}}

   \approx 1.80\times10^{4}\ \text{N C}^{-1}

   \]

4. Direction: toward the charge, i.e. \(-\hat z\).
5. Vector result:

   \[

   \boxed{\displaystyle \vec E = -1.80\times10^{4}\ \hat z\ \text{N C}^{-1}}

   \]

Superposition of Electric Fields (Syllabus 8)

When more than one point charge is present, the total field at a point is the vector sum of the individual fields:

\[

\vec E{\text{total}} = \sum{i}\frac{Qi}{4\pi\varepsilon0 ri^{2}}\;\hat ri

\]

Key points for superposition:

• Calculate each field separately (magnitude + direction).
• Resolve vectors into components (usually Cartesian) before adding.
• Combine the components to obtain the resultant magnitude and direction.

Uniform Electric Fields (Syllabus 18.2) – Sidebar

Between two large, parallel conducting plates with a potential difference \(\Delta V\) and separation \(\Delta d\), the field is approximately uniform:

\[

\vec E = \frac{\Delta V}{\Delta d}\,\hat n

\]

Example: \(\Delta V = 12\ \text{V}\), \(\Delta d = 3.0\ \text{cm}\) → \(|\vec E| = 4.0\times10^{2}\ \text{V m}^{-1}\) directed from the negative to the positive plate.

Link to Electric Potential (Syllabus 18.5)

The electric potential of a point charge is

\[

V = \frac{Q}{4\pi\varepsilon_0 r}\qquad\text{(reference }V=0\text{ at }r\to\infty\text{)}

\]

Because \(\vec E = -\nabla V\), for a purely radial field

\[

E = -\frac{dV}{dr}= \frac{Q}{4\pi\varepsilon_0 r^{2}}

\]

This shows that mastering the field formula automatically gives the corresponding potential.

Design an Experiment – Mapping the Field of a Point Charge (AO3)

Common Mistakes to Avoid

• Forgetting to square the distance \(r\).
• Using centimetres or other non‑SI units for \(r\); always convert to metres.
• Omitting the factor \(4\pi\) (the result becomes ≈ 12.57 times too large).
• Confusing the sign of \(Q\) with the direction of \(\vec E\); the sign only tells you whether \(\hat r\) points outward or inward.
• Leaving out the unit vector \(\hat r\) when a vector answer is required.
• Adding magnitudes instead of vectors when applying superposition.

Concept‑Link Table

Quick Reference Table

Suggested Diagram

Summary

The electric field produced by an isolated point charge in free space follows the inverse‑square law:

\[

\boxed{\displaystyle \vec E = \frac{Q}{4\pi\varepsilon_0\,r^{2}}\;\hat r}

\]

Remember:

• Square the distance and keep it in metres.
• Include the constant \(4\pi\varepsilon_0\) (≈ 1.112 × 10\(^{-10}\) C\(^2\) N\(^{-1}\) m\(^{-2}\)).
• Direction is given by \(\hat r\); outward for \(Q>0\), inward for \(Q<0\).
• Use superposition to combine fields from several charges.
• The same expression underpins the relationship between electric field and electric potential and provides the basis for experimental field‑mapping.

Further Reading (Next Syllabus Sections)

• 18.5 – Electric potential and equipotential surfaces.
• 19 – Capacitance and energy stored in electric fields.
• 20 – Magnetic fields and electromagnetic induction.
• 21 – Alternating current, transformers and power.
• 22 – Quantum and nuclear physics (applications of electric fields).