U3AOS2 Topic 4: Electrical Fields

A region where a force would be applied on other charged particles that enter it is known as an electric field. Since it has both magnitude and direction, it is a vector field. Michael Faraday was the one who first proposed the idea of the electric field.


An electric field's strength is defined as the force per unit charge exerted on a positive charge placed at any point in the field. Mathematically, it is modelled by: 

\[ E= \frac{F}{q} \] 

where:


  1. E is the electric field strength in $ N / C $,
  2. F is the force in $ N $ experienced by a charge within the field,
  3. q is the magnitude of the test charge in Columbs ($ C $).


It's important that we're using the magnitude of the charge, as charge is a vector quantity it can be negative in physics, where it would be a negative charge.


Electrical fields are from the perspective of positive charges. In electrical fields, like charges repel. This can be seen in the simulation below.


Fields do not enter the particles. The arrows entering the particles is a bug in the simulation, when modelling the forces the should get close but not touch the particle(s).


Behaviour of Electric Fields


The formula for electrical field strength is identical to the gravitational field strength formula however the difference is the universal gravitation constant G is replaced with k and the mass being replaced with charge denoted as q.


\[ E = k\frac{q}{r^2} \]​

where $ k=8.99 \times 10^9 $ is the Coloumb's constant in $ \frac{N m^2}{C^2} $



The direction of the electric field is radially outward from a positive charge and radially inward toward a negative charge. The strength of the electric field decreases with the square of the distance from the charge. The inverse-square law indicates that if you double the distance from a charge, the electric field strength becomes one-fourth as strong.


Uniform electric fields


The field above is not uniform, which means the strength of the field changes depending on how far you are from it. However, this is not always the case.


In uniform electric fields (generally between charged plates), the field strength will NOT change and is calculated by:

\[ E = \frac{V}{d} \]

where V is the voltage from the plates in volts and d is the distance between the plates in $ m $.



Note that the electric field strength here is in $ V/m $, which is equivalent to $ N/C $, it's a different unit to represent the same quantity and can be used for any electric field.


Sources of Electric Fields


Electric fields are generated by electric charges. There are two main types of electric charges that affect the direction of the applied electric force:

  1. Positive charges, which produce an outward electric field.
  2. Negative charges, which produce an inward electric field.


Electric fields can also be produced by multiple charges interacting with each other. The resultant electric field at a point due to multiple charges is the vector sum of the fields due to individual charges, known as the principle of superposition.


These fields are visualized using electric field lines:

  1. Electric field lines point away from positive charges and toward negative charges.
  2. The density of the lines indicates the strength of the field. closer lines represent stronger fields.


Electric Field and Potential


The electric field is related to the electric potential $ V $ by the negative gradient:

\[ E = − \nabla V \]


This relationship models the fact that the electric field points in the direction of the greatest decrease of electric potential. The potential difference between two points in an electric field provides a measure of the work done to move a charge between those points.


Electric Field Inside a Conductor


In electrostatic equilibrium, the electric field inside a conductor is zero. Charges in a conductor redistribute themselves on the surface to cancel any internal electric field. This leads to the concept of shielding, where the interior of a conductor can be protected from external electric fields.


Applications of electric fields


Capacitors: Devices that store electric energy by maintaining a potential difference between two conductors. The electric field between the conductors determines the capacitor's behavior.

Electrostatic Precipitators: Used in industrial processes to remove particulate matter from exhaust gasses using electric fields.

Medical Devices: Techniques like electrocardiography (ECG) use electric fields to measure heart activity.


Electric fields are also fundamental to the functioning of many electronic devices, such as transistors and sensors.

Example 1

 What is the magnitude of the electric field at a distance of 25 cm from a point charge of 5 µC?

Using E=k⋅∣q∣r2E = \frac{k \cdot |q|}{r^2}
  • k=8.99×109 N m2/C2k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2
  • q=5×10−6 Cq = 5 \times 10^{-6} \, \text{C}
  • r=0.25 mr = 0.25 \, \text{m}

E=(0.25)28.99×109⋅5×10−6​


=2.88×106N/C

Example 2

 Calculate the electric field at a point 0.5 m away from an infinitely long line charge with a linear charge density of 2×10−9 C/m2 \times 10^{-9} \, \text{C/m}.

Using E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}
  • λ=2×10−9 C/m\lambda = 2 \times 10^{-9} \, \text{C/m}
  • ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}
  • r=0.5 mr = 0.5 \, \text{m}

​E=2π⋅8.85×10−12⋅0.52×10−9​=113.4N/C

Example 3

Find the electric field on the axial line of a dipole consisting of charges +q+q and −q-q, separated by 2 cm.

Using E=14πϵ0⋅2qd(r2+d2)3/2E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2qd}{(r^2 + d^2)^{3/2}}
  • q=1×10−6 Cq = 1 \times 10^{-6} \, \text{C}
  • d=0.02 md = 0.02 \, \text{m}
  • r=0.1 mr = 0.1 \, \text{m}

E=4π⋅8.85×10−121​⋅(0.12+0.022)3/22⋅1×10−6⋅0.02​

≈2.8×103N/C

Example 4

 What is the magnitude of the electric field at a distance of 25 cm from a point charge of 5 µC?

Using E=k⋅∣q∣r2E = \frac{k \cdot |q|}{r^2}
  • k=8.99×109 N m2/C2k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2
  • q=5×10−6 Cq = 5 \times 10^{-6} \, \text{C}
  • r=0.25 mr = 0.25 \, \text{m}

E=(0.25)28.99×109⋅5×10−6​


=2.88×106N/C

Example 5

 Calculate the electric field at a point 0.5 m away from an infinitely long line charge with a linear charge density of 2×10−9 C/m2 \times 10^{-9} \, \text{C/m}.

Using E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}
  • λ=2×10−9 C/m\lambda = 2 \times 10^{-9} \, \text{C/m}
  • ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}
  • r=0.5 mr = 0.5 \, \text{m}

​E=2π⋅8.85×10−12⋅0.52×10−9​=113.4N/C

Example 6

Find the electric field on the axial line of a dipole consisting of charges +q+q and −q-q, separated by 2 cm.

Using E=14πϵ0⋅2qd(r2+d2)3/2E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2qd}{(r^2 + d^2)^{3/2}}
  • q=1×10−6 Cq = 1 \times 10^{-6} \, \text{C}
  • d=0.02 md = 0.02 \, \text{m}
  • r=0.1 mr = 0.1 \, \text{m}

E=4π⋅8.85×10−121​⋅(0.12+0.022)3/22⋅1×10−6⋅0.02​

≈2.8×103N/C

Example 7

 What is the magnitude of the electric field at a distance of 25 cm from a point charge of 5 µC?

Using E=k⋅∣q∣r2E = \frac{k \cdot |q|}{r^2}
  • k=8.99×109 N m2/C2k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2
  • q=5×10−6 Cq = 5 \times 10^{-6} \, \text{C}
  • r=0.25 mr = 0.25 \, \text{m}

E=(0.25)28.99×109⋅5×10−6​


=2.88×106N/C

Example 8

 Calculate the electric field at a point 0.5 m away from an infinitely long line charge with a linear charge density of 2×10−9 C/m2 \times 10^{-9} \, \text{C/m}.

Using E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}
  • λ=2×10−9 C/m\lambda = 2 \times 10^{-9} \, \text{C/m}
  • ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}
  • r=0.5 mr = 0.5 \, \text{m}

​E=2π⋅8.85×10−12⋅0.52×10−9​=113.4N/C

Example 9

Find the electric field on the axial line of a dipole consisting of charges +q+q and −q-q, separated by 2 cm.

Using E=14πϵ0⋅2qd(r2+d2)3/2E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2qd}{(r^2 + d^2)^{3/2}}
  • q=1×10−6 Cq = 1 \times 10^{-6} \, \text{C}
  • d=0.02 md = 0.02 \, \text{m}
  • r=0.1 mr = 0.1 \, \text{m}

E=4π⋅8.85×10−121​⋅(0.12+0.022)3/22⋅1×10−6⋅0.02​

≈2.8×103N/C

Exercise 1

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