Electric field strength (2.0) Revision Notes for AQA A-Level Physics

Electric field strength (2.0)

Electric field strength is a measure of the force experienced by a charged particle placed within an electric field. Understanding how to define and calculate field strength is essential for analyzing the behaviour of charges in various field configurations.

Defining electric field strength

Electric field strength at a point describes how strong the electric field is at that location. The definition is based on the force experienced by a test charge.

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Electric field strength E at a point in an electric field is defined as the force per unit charge acting on a small positive test charge placed at that point.

The defining equation is:

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

where:

E is the electric field strength (NC⁻¹)
F is the force on the test charge (N)
Q is the magnitude of the test charge (C)

Electric field strength is a vector quantity. Its direction at any point is the direction of the force that would act on a positive test charge placed at that point.

The test charge

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The concept of a test charge is central to defining field strength. A test charge is a small positive charge used to probe the electric field without significantly altering it.

The test charge must satisfy two conditions:

It must be sufficiently small in magnitude so that its own electric field does not distort the field being measured
It must be small enough that it does not affect the distribution of charges creating the field

In practice, we imagine placing a vanishingly small positive charge at a point to determine the field strength there.

Electric field strength in radial fields

A radial field exists around an isolated point charge or a uniformly charged sphere. For a uniformly charged sphere, when applying Coulomb's law, the charge can be treated as if it were concentrated at the centre of the sphere.

The electric field strength at a distance r from a point charge Q is determined using Coulomb's law. The force between the charge Q and a small test charge q placed at distance r is:

$F = \frac{Qq}{4\pi\varepsilon_0 r^2}$

where ε₀ is the permittivity of free space.

Since E = F/q, the electric field strength due to charge Q at distance r is:

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

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This equation shows that in a radial field:

The field strength decreases with the square of the distance from the charge
The field is stronger closer to the charge
The field strength depends only on the magnitude of the charge Q and the distance r

The permittivity of free space, ε₀, is a constant with the value:

$\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$

Electric field strength in uniform fields

A uniform electric field exists between two parallel metal plates when a potential difference is applied across them. In a uniform field, the electric field strength has the same magnitude and direction at all points between the plates.

For a uniform field, the relationship between electric field strength E, potential difference V, and separation d between the plates is:

$E = \frac{V}{d}$

where:

E is the electric field strength (Vm⁻¹ or NC⁻¹)
V is the potential difference between the plates (V)
d is the separation between the plates (m)

This equation can also be written as:

$E = \frac{\Delta V}{d}$

where ΔV represents the potential difference.

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In a uniform field:

A charged particle experiences the same force regardless of its position between the plates
The field lines are equally spaced and parallel
The field strength remains constant throughout the region

Units of electric field strength

From the defining equation E = F/Q, the unit of electric field strength is newton per coulomb (NC⁻¹).

From the uniform field equation E = V/d, the unit can also be expressed as volt per metre (Vm⁻¹).

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These two units are equivalent:

$1 \text{ NC}^{-1} = 1 \text{ Vm}^{-1}$

Both units are acceptable and widely used in physics. The choice often depends on context: NC⁻¹ emphasizes the force aspect, while Vm⁻¹ emphasizes the potential gradient aspect.

Worked example: calculating electric field strength

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Worked Example: Electric field strength near a gold nucleus

Problem: Determine the electric field strength at a distance of 50 fm from a gold nucleus. The proton number of gold is 79, and the permittivity of free space is 8.85 × 10⁻¹² Fm⁻¹.

Solution:

Step 1: Calculate the charge on the gold nucleus

The charge on the gold nucleus is Q = 79e, where e = 1.6 × 10⁻¹⁹ C.

Therefore: Q = 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C

Step 2: Convert the distance to metres

Distance from nucleus: r = 50 fm = 50 × 10⁻¹⁵ m = 5.0 × 10⁻¹⁴ m

Step 3: Apply the radial field equation

Using the equation for a radial field:

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

$E = \frac{1.264 \times 10^{-17}}{4\pi \times 8.85 \times 10^{-12} \times (5.0 \times 10^{-14})^2}$

$E = \frac{1.264 \times 10^{-17}}{4\pi \times 8.85 \times 10^{-12} \times 2.5 \times 10^{-27}}$

$E = 4.5 \times 10^{21} \text{ NC}^{-1}$

Answer: The electric field strength is 4.5 × 10²¹ NC⁻¹

This extremely large field strength is typical near atomic nuclei.

Remember!

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Key Points to Remember:

Electric field strength E is defined as force per unit charge: E = F/Q
The unit of E can be expressed as either NC⁻¹ or Vm⁻¹ (both are equivalent)
For radial fields around point charges: E = Q/(4πε₀r²), where field strength decreases with the square of distance
For uniform fields between parallel plates: E = V/d, where field strength is constant throughout
A test charge must be small enough not to disturb the field being measured

Electric field strength (2.0) (AQA A-Level Physics): Revision Notes