U3AOS2 Topic 4: Electrical Fields

A region where a force would be applied on other charged particles 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 is defined as the force per unit charge exerted on a positive test charge placed at a point in the field. Mathematically, it is given by: 

                                             E=F/q 

where,

  • E is the electric field,

  • F is the force experienced by the test charge,

  • q is the magnitude of the test 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 in real life they would get close but not touch.

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


E=kqr2E = k\frac{q}{r^2} where k=8.99×109k=8.99 \times 10^9




Explanation

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. This inverse-square law indicates that if you double the distance from a charge, the electric field strength becomes one-fourth as strong.

Electric fields can also be produced by multiple charges. 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.

In practical applications, electric fields are crucial in understanding phenomena in electrostatics and electrodynamics. For example, in capacitors, an electric field is created between two plates with opposite charges, storing energy. Electric fields are also fundamental to the functioning of many electronic devices, such as transistors and sensors.


The field above is not uniform meaning 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 dividing the voltage by the distance between the plates.


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



Units

The SI unit of the electric field is volts per meter (V/m), which is equivalent to newtons per coulomb (N/C).

Understanding electric fields is essential for studying various aspects of physics and engineering, as they describe how charged particles interact with each other and with electric fields, leading to the development of various technologies and scientific theories.

Sources of Electric Fields

Electric fields are generated by electric charges. There are two main types of electric charges:

  1. Positive charges, which produce an outward electric field.

  2. Negative charges, which produce an inward electric field.

These fields can be visualized using electric field lines:

  • Electric field lines point away from positive charges and toward negative charges.

  • 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 VVV by the negative gradient:

    E=−∇V

    This relationship implies 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:

    • 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.


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&& (&&1&& Question)

 Determine the electric field on the axis of a uniformly charged disk of radius 0.1 m with a surface charge density σ=5×10−6 C/m2\sigma = 5 \times 10^{-6} \, \text{C/m}^2 at a point 0.2 m from its center.

Submit

Exercise &&2&& (&&1&& Question)
Find the electric field at a point 0.15 m outside a uniformly charged sphere of radius 0.1 m with a total charge of 1×10−6 C1 \times 10^{-6} \, \text{C}.
Submit

Exercise &&3&& (&&1&& Question)
 Calculate the electric field at a point 0.05 m inside a sphere of radius 0.1 m with a charge of 1×10−6 C1 \times 10^{-6} \, \text{C}.
Submit

Exercise &&4&& (&&1&& Question)

Find the electric field at the midpoint between two charges, +2μC+2 \mu\text{C} and −2μC-2 \mu\text{C}, placed 0.1 m apart.

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