Electric field strength

CIE A-Level Physics· 18.1 Electric fields and field strength· 15 min read

1. Definition of Electric Field Strength★★☆☆☆⏱ 5 min

📘 Definition

Electric field strength

E

Force per unit positive test charge placed at a point in an electric field

Example:

A 1 C test charge experiences 5 N force, so E = 5 N C⁻¹ at that point

Electric field strength is a vector quantity, so it has both magnitude and direction. The direction of E matches the direction of force that acts on a positive test charge placed at the point.

E = \frac{F}{q}

Where $F$ is the force acting on the test charge, and $q$ is the magnitude of the test charge.

📐 Worked Example

A test charge of $+ 2.0 \times 1 0^{- 6}$ C experiences an electrostatic force of 0.04 N in an electric field. Calculate the electric field strength at that point.

1
Recall the definition formula for electric field strength:
2
$E = \frac{F}{q}$
3
Substitute the given values for force and test charge:
4
$E = \frac{0.04}{2.0 \times 10^{-6}} = 2.0 \times 10^4 \text{ N C}^{-1}$
5
Direction of E matches the direction of force acting on the positive test charge.

Exam tip:

Always remember E is defined per positive test charge for direction conventions

2. Uniform and Radial Field Strength★★★☆☆⏱ 6 min

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The two most common field configurations you will solve problems for are uniform fields between parallel charged plates, and radial fields around point charges or charged spheres.

📘 Definition

Uniform electric field

A field where electric field strength has the same magnitude and direction at all points (away from plate edges)

Example:

Between two oppositely charged parallel plates connected to a fixed potential difference

E = \frac{V}{d}

Where $V$ is the potential difference between the plates, and $d$ is the perpendicular separation of the plates.

For a radial field around a point source charge, we derive field strength from Coulomb's law, giving the inverse square relation:

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

Where $Q$ is the source charge creating the field, $r$ is the distance from the source, and $ε_{0}$ is the permittivity of free space.

📐 Worked Example

Two parallel plates are separated by 5.0 cm, with a potential difference of 100 V between them. Calculate the electric field strength between the plates.

1
Convert plate separation to SI units (metres):
2
$d = 5.0 \text{ cm} = 0.050 \text{ m}$
3
Use the uniform field formula:
4
$E = \frac{V}{d} = \frac{100}{0.050} = 2000 \text{ N C}^{-1}$

📐 Worked Example

Calculate the electric field strength at 0.1 m from a point charge of $+ 1.0 \times 1 0^{- 6}$ C. ( $ε_{0} = 8.85 \times 1 0^{- 12}$ F m⁻¹)

1
Substitute into the radial field formula:
2
$E = \frac{Q}{4 \pi \varepsilon_0 r^2} = \frac{1.0 \times 10^{-6}}{4 \pi (8.85 \times 10^{-12}) (0.1)^2} = 9.0 \times 10^5 \text{ N C}^{-1}$

Exam tip:

Always convert distance units to metres before substituting into formulas

3. Field Strength as Potential Gradient★★★☆☆⏱ 4 min

Electric field strength can also be defined in terms of electric potential. The magnitude of electric field strength at any point equals the negative potential gradient at that point.

E = - \frac{dV}{dr}

The negative sign indicates that electric field always points in the direction of decreasing electric potential. For uniform fields, this simplifies to the familiar $E = V / d$ , since potential changes at a constant rate.

📐 Worked Example

The potential at 2 mm from a point charge is 200 V, and at 3 mm it is 133 V. Estimate the electric field strength at this position.

1
Approximate the derivative $d V / d r$ with the finite difference $Δ V /Δ r$ :
2
$\Delta V = 200 - 133 = 67 \text{ V}, \Delta r = 1 \times 10^{-3} \text{ m}$
3
Take the magnitude to get E:
4
$E = \left| \frac{\Delta V}{\Delta r} \right| = \frac{67}{1 \times 10^{-3}} = 6.7 \times 10^4 \text{ N C}^{-1}$

4. Common Pitfalls

Wrong move:

Confusing test charge and source charge in definitions and formulas

Why:

E is defined per test charge, but radial E uses the source charge creating the field

Correct move:

Remember: definition $E = F / q$ (q = test charge), radial $E = Q / (4 π ε_{0} r^{2})$ (Q = source charge)

Wrong move:

Leaving plate separation in centimetres when calculating $E = V / d$

Why:

SI units for distance are metres, so this gives an incorrect magnitude for E

Correct move:

Always convert all distances to metres before substituting into any field formula

Wrong move:

Assuming E = 0 at a point where potential V = 0

Why:

Potential is a scalar, E is the gradient of potential; zero potential does not mean zero gradient

Correct move:

Calculate E independently from potential, never assume one is zero if the other is

Wrong move:

Using the radial inverse square formula for points inside a charged conducting sphere

Why:

No charge is enclosed inside a conducting sphere, so the field is zero inside

Correct move:

Use E = 0 for all points inside a charged conducting sphere, only use the radial formula for points outside

5. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Definition of E	$E = \frac{F}{q}$	Force per positive test charge, vector
Uniform field (plates)	$E = \frac{V}{d}$	Constant magnitude and direction
Radial field (point charge)	$E = \frac{Q}{4 π ε _{0} r ^{2}}$	Inverse square law, $E \propto 1/ r^{2}$
Potential gradient	$E = - \frac{d V}{d r}$	E points to lower potential
Charged conducting sphere	$E = 0$ (inside), $E = \frac{Q}{4 π ε _{0} r ^{2}}$ (outside)	Same as point charge outside

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 22
Uniform plate field strength calculation
2023 · 12
Radial field strength comparison
2021 · 21
Potential gradient relation to E

Going deeper

syllabusCIE 9702 A-Level Physics SyllabusSee section 18 for full electrostatics requirements

What's Next

Now that you have mastered electric field strength, you can move on to related concepts that build directly on this foundation. Calculating E is essential for solving problems involving motion of charged particles in electric fields, a common long exam question. It also underpins the study of capacitance, where you will use the uniform field relation to derive capacitance of parallel plate capacitors. The link between E and potential gradient also connects directly to equipotential surfaces, which are frequently tested alongside field strength. Mastery of this sub-topic makes all subsequent electrostatics topics far easier to understand.