# Electric field strength

> CIE A-Level Physics · 9702 A2
> Source: https://www.owlsprep.com/study/cie-9702-u21-electric-field-strength/

This sub-topic covers the definition of electric field strength, calculations for uniform and radial fields, and the relationship between field strength and potential gradient. It is a foundational concept for all electrostatics problems in A2 Physics.

**Prerequisites:** [Coulomb's Law](https://www.owlsprep.com/study/cie-9702-u21-coulombs-law/); [Electric Potential](https://www.owlsprep.com/study/cie-9702-u21-electric-potential/)

## Learning objectives

- Define electric field strength correctly for all field configurations
- Calculate field strength for uniform parallel plate and radial point charge fields
- Relate electric field strength to potential gradient
- Apply the inverse square law to radial electric fields

## Definition of Electric Field Strength

**Electric field strength** — Force per unit positive test charge placed at a point in an electric field

*Notation:* E

*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 $+2.0 \times 10^{-6}$ 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

## Uniform and Radial Field Strength

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.

**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 $\varepsilon_0$ is the permittivity of free space.

> **info**
>
> For a charged conducting sphere, the field outside the sphere is identical to that of a point charge at the sphere's centre. The field inside the sphere is zero.

**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 $+1.0 \times 10^{-6}$ C. ($\varepsilon_0 = 8.85 \times 10^{-12}$ 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

*Calculator:* allowed

## Field Strength as Potential Gradient

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.

> **note**
>
> In CIE exams, you only need to give the magnitude of potential gradient for E; the negative sign is only required if you are asked for direction.

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

## Common pitfalls

- **Wrong:** Confusing test charge and source charge in definitions and formulas
  - Why it fails: E is defined per test charge, but radial E uses the source charge creating the field
  - Correct: Remember: definition $E = F/q$ (q = test charge), radial $E = Q/(4πε₀r²)$ (Q = source charge)
- **Wrong:** Leaving plate separation in centimetres when calculating $E = V/d$
  - Why it fails: SI units for distance are metres, so this gives an incorrect magnitude for E
  - Correct: Always convert all distances to metres before substituting into any field formula
- **Wrong:** Assuming E = 0 at a point where potential V = 0
  - Why it fails: Potential is a scalar, E is the gradient of potential; zero potential does not mean zero gradient
  - Correct: Calculate E independently from potential, never assume one is zero if the other is
- **Wrong:** Using the radial inverse square formula for points inside a charged conducting sphere
  - Why it fails: No charge is enclosed inside a conducting sphere, so the field is zero inside
  - Correct: Use E = 0 for all points inside a charged conducting sphere, only use the radial formula for points outside

## 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) | $E = \frac{Q}{4 \pi \varepsilon_0 r^2}$ | Inverse square law, $E \propto 1/r^2$ |
| Potential gradient | $E = -\frac{dV}{dr}$ | E points to lower potential |
| Charged conducting sphere | $E=0$ (inside), $E = \frac{Q}{4 \pi \varepsilon_0 r^2}$ (outside) | Same as point charge outside |

## 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.

- [Electric potential](https://www.owlsprep.com/study/cie-9702-u21-electric-potential/)
- [Charged particle motion in E-fields](https://www.owlsprep.com/study/cie-9702-u21-charged-particle-motion-in-e/)
- [Capacitance](https://www.owlsprep.com/study/cie-9702-u22-overview/)

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