How Do You Calculate Electric Field Strength (E = F/q)?

For both SL and HL, calculate electric field strength by dividing the electric force on a small positive test charge by its charge: $E = \frac{F}{q}$. Express the magnitude in newtons per coulomb, $\text{N C}^{-1}$, and state the field direction when required.

The Reasoning and Mechanism

Electric field strength at a point is defined as the electric force per unit positive charge placed at that point:

$$E = \frac{F}{q}$$

Here, $E$ is electric field strength in $\text{N C}^{-1}$, $F$ is electric force in newtons, and $q$ is the test charge in coulombs. This relationship appears in IB Physics subtopic D.2 Electric and magnetic fields and applies at both SL and HL.

To calculate reliably, first convert the charge to coulombs. Next, substitute the force magnitude and charge magnitude into the equation. Finally, determine direction separately from the sign of the charge. The equation gives the magnitude when positive magnitudes are substituted; it does not by itself encode the vector direction.

Electric field strength is a vector quantity. Its direction is defined as the direction of the force that would act on a positive test charge. A positive charge experiences force in the field direction, while a negative charge experiences force in the opposite direction.

Worked Example

A positive test charge of $2.0 \times 10^{-6}\,\text{C}$ experiences an electric force of $0.060\,\text{N}$. Calculate the field strength.

$$E = \frac{F}{q} = \frac{0.060}{2.0 \times 10^{-6}} = 3.0 \times 10^4\,\text{N C}^{-1}$$

Therefore, the electric field strength is $3.0 \times 10^4\,\text{N C}^{-1}$, in the same direction as the force because the test charge is positive.

A common misconception is that a larger test charge creates a larger electric field. The field is determined by the source charges; increasing $q$ increases $F$ proportionally, so (F/q) remains constant, provided the test charge does not significantly disturb the source charges.

IB Exam Technique

For a Calculate question, show the equation, substitution, and final value. Use the charge in coulombs, state $\text{N C}^{-1}$, and include a direction when asked; for a negative charge, the force and field directions are opposite.