3.7.3.2
Electric field strength is force per unit charge
Electric Fields & Potential — AQA A-Level Physics
Key Definition
Electric field strength — The force per unit charge experienced by a small positive test charge placed at that point.
$$\begin{aligned}
E &= \frac{F}{Q} \\
&= \frac{V}{d}
\end{aligned}$$
$$E = \frac{F}{Q}$$
- $E$: electric field strengthThe force per unit positive chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). at a point in an electric field. Measured in N C⁻¹ or V m⁻¹. (N C^-1 or V m^-1)
- $F$: electric force on the chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). (N)
- $Q$: magnitude of the chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). (C)
- Electric field strengthThe force per unit positive charge at a point in an electric field. Measured in N C⁻¹ or V m⁻¹. is a vector quantityA quantity that has both magnitude and direction..
- Direction: away from a positive charge, towards a negative charge.
- A positive charge +Q experiences a force EQ in the direction of the field.
- A negative charge -Q experiences a force EQ in the opposite direction to the field.
- N C^-1 and V m^-1 are equivalent units for E.
Worked Example
A charged particle experiences a force of 0.3 N at a point where the electric field strengthThe force per unit positive charge at a point in an electric field. Measured in N C⁻¹ or V m⁻¹. is \(3.5 \times 10^{4}\) N C^-1. Calculate the magnitude of the charge.
Show Solution
1
Write the equation
$$E = \frac{F}{Q}$$
2
Rearrange for Q
$$Q = \frac{F}{E}$$
3
Substitute and calculate
$$Q = \frac{0.3}{3.5 \times 10^{4}} = 8.6 \times 10^{-6} \text{ C (2 s.f.)}$$
Answer
$Q = 8.6 \; \mu\text{C}$