D.2 Electric fields

IB Physics HL· 30 min read

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

An electric field is a region of space where a stationary electric charge experiences a force. It describes how the electrostatic force acts on charges at any point in the region, and follows the principle of superposition for multiple sources.

📘 Definition

Electric field strength

E

Force per unit positive test charge at a point, defined as $E = \frac{F}{q}$ , where $F$ is the force on test charge $q$ . Units are newtons per coulomb (N C⁻¹) or equivalent volts per metre (V m⁻¹).

📐 Worked Example

A test charge of $+ 1.5 \times 1 0^{- 6}$ C experiences an upward electrostatic force of $4.5 \times 1 0^{- 3}$ N in an electric field. Calculate the magnitude and direction of the field strength.

1
Use the definition of electric field strength:
2
$E = \frac{F}{q} = \frac{4.5 \times 10^{-3}\ \text{N}}{1.5 \times 10^{-6}\ \text{C}} = 3000\ \text{N C}^{-1}$
3
By definition, electric field direction matches the direction of force on a positive test charge, so the field points upwards.

2. Radial Electric Fields from Point Charges★★★☆☆⏱ 8 min

A single point charge $Q$ produces a radial electric field, where field strength decreases with the square of distance from the charge. This formula is derived directly from Coulomb's law: $F = \frac{k Qq}{r ^{2}}$ , so dividing by $q$ gives the field strength.

📐 Worked Example

Calculate the electric field strength 0.10 m away from a point charge of $- 2.0 \times 1 0^{- 6}$ C, given $k = 9.0 \times 1 0^{9}$ N m² C⁻².

1
Use the radial field formula:
2
$E = \frac{kQ}{r^2} = \frac{(9.0 \times 10^9)(-2.0 \times 10^{-6})}{(0.10)^2}$
3
$E = -1.8 \times 10^6\ \text{N C}^{-1}$
4
The negative sign indicates the field points towards the negative charge, with magnitude $1.8 \times 1 0^{6}$ N C⁻¹.

3. Uniform Electric Fields Between Parallel Plates★★☆☆☆⏱ 6 min

When two parallel conducting plates are connected to a constant potential difference $V$ , a nearly uniform electric field forms between the plates (edge effects are ignored for most exam problems). Field strength is constant across all points between the plates.

📘 Definition

Uniform field strength

E

For parallel plates separated by distance $d$ , field strength is given by $E = \frac{V}{d}$ , where $V$ is the potential difference between the plates.

📐 Worked Example

Two parallel plates separated by 5.0 mm are connected to a 12 V battery. Calculate the electric field strength between the plates.

1
Convert separation to SI units: $d = 5.0$ mm $= 5.0 \times 1 0^{- 3}$ m
2
Apply the uniform field formula:
3
$E = \frac{V}{d} = \frac{12\ \text{V}}{5.0 \times 10^{-3}\ \text{m}} = 2400\ \text{V m}^{-1}$
4
Since $1$ V m⁻¹ $= 1$ N C⁻¹, the field strength is 2400 N C⁻¹, pointing from the positive plate to the negative plate.

4. Motion of Charged Particles in Uniform Fields★★★★☆⏱ 10 min

A charged particle in a uniform electric field experiences a constant force $F = q E$ , so it accelerates at a constant rate $a = \frac{q E}{m}$ , where $m$ is the particle mass. When a particle enters the field perpendicular to the field lines, it follows a parabolic projectile path, just like motion in a gravitational field.

📐 Worked Example

An electron with mass $m_{e} = 9.1 \times 1 0^{- 31}$ kg and charge $e = 1.6 \times 1 0^{- 19}$ C is in a uniform electric field of 200 N C⁻¹ directed downwards. Calculate the acceleration of the electron.

1
Calculate force on the electron: $F = e E$ . The electron is negative, so force acts opposite to the field direction (upwards).
2
Use Newton's second law $a = \frac{F}{m _{e}}$ :
3
$a = \frac{eE}{m_e} = \frac{(1.6 \times 10^{-19}\ \text{C})(200\ \text{N C}^{-1})}{9.1 \times 10^{-31}\ \text{kg}} \approx 3.5 \times 10^{13}\ \text{m s}^{-2}$
4
Acceleration is directed upwards, opposite to the field direction.

5. Common Pitfalls

Wrong move:

Treating electric field strength as a scalar, adding magnitudes regardless of direction for multiple charges.

Why:

Electric field is a vector quantity, so direction determines whether fields add or cancel.

Correct move:

Assign directions to each field based on source charge sign, then add as vectors.

Wrong move:

Using $E = V / d$ for radial point charge fields, or $E = k Q / r^{2}$ for uniform parallel plate fields.

Why:

Each formula only applies to its specific type of electric field.

Correct move:

First identify the type of field, then use the matching formula.

Wrong move:

Taking electric field direction as the direction of force on a negative charge.

Why:

The definition of electric field uses a positive test charge by convention.

Correct move:

Reverse the direction of force if the charge is negative when finding field direction.

Wrong move:

Forgetting that the sign of the field value indicates direction, not magnitude.

Why:

Negative values come from negative source charges, they do not mean the field is smaller.

Correct move:

Report magnitude as an absolute value, and state direction separately if required.

6. Quick Reference Cheatsheet

Concept	Formula	Key Direction Rule
Field strength (definition)	$E = \frac{F}{q}$	Same as force on +ve test charge
Radial field (point charge)	$E = \frac{k Q}{r ^{2}} = \frac{Q}{4 π ε _{0} r ^{2}}$	Away from +Q, towards -Q
Uniform field (parallel plates)	$E = \frac{V}{d}$	From +ve plate to -ve plate
Force on charge	$F = q E$	Same as E for +ve charge, opposite for -ve charge

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.

2023 · 1
Point charge field strength calculation
2022 · 2
Electron motion in uniform field
2021 · 1
Uniform field between parallel plates

What's Next

Mastering electric fields is critical for understanding electric potential, energy changes in electrostatics, and Gauss's law for IB Physics HL. This topic also forms the foundation for analysing combined electric and magnetic fields, which are used to explain phenomena like particle deflection and electromagnetic induction. Strong skills in field calculations will make more advanced topics in the fields theme significantly easier to master.