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Introduction to Electric Fields

 

This topic is part of the HSC Physics course under the section Electrostatics.

HSC Physics Syllabus

  • using the electric field lines representation, model qualitatively the direction and strength of electric fields produced by:

    – simple point charges
    – pairs of charges
    – dipoles
    – parallel charged plates
  • apply the electric field model to account for and quantitatively analyse interactions between charged objects using: 
– `F=qE` (ACSPH103, ACSPH104) 
– `E=Vd`

Introduction to Electric Fields

    Electric Fields

    An electric field is a region around a charged particle or charged object where its electric force is exerted on other charges. It can be defined as the amount of force exerted on a charge per unit of its charge.

    The field can be visualised using field lines:

    • Lines originate from positive charges and end on negative charges.
    • The strength of the field is indicated by the density of the lines. When field lines are drawn closer together, the electric field strength is greater.
    • Direction of the field is the direction a positive charge would move if placed in the electric field.
    • The unit of electric field is Newton per Coulomb (N C–1) or volt per metre (V m–1)

    Simple Point Charges

    Electric field of point charges

     

    A single point charge generates an electric field radiating outward if positive or inward if negative. The field lines are radial, indicating the field's strength is the same at equal distances from the charge. However, as the distance increase, the field strength decreases.

    The strength of the field `E` is given by Coulomb’s law:

    where

    • `q` is the charge (unit: C)
    • `r` is the distance from the charge (unit: m)
    • `\varepsilon_0` is the electric permittivity of free space (unit: A2 s4 kg-1 m-3)

     

    Coulomb's Law highlights that the field strength is directly proportional to the charge that produces the field, and inversely proportional to the distance from the charge squared. As seen in the above diagrams, the field line density and therefore field strength decreases with increasing distance from the charge.

    Pairs of Charges

    Diagram shows an electric dipole: two equal and opposite charges

     

    For a pair of opposite charges, the field lines begin on the positive charge and end on the negative charge. An electric dipole consists of two equal and opposite charges in close proximity. The field lines start on the positive charge and end on the negative charge, creating a pattern that extends outwards and loops back from the negative to the positive charge. The strength of the field in the mid-region between the charges is strong and decreases with distance.

     

     

    When the charges are like, the field lines repel each other, indicating that like charges repel. The electric field strength at the midpoint between two like charges of equal magnitude is exactly zero. This is because the electric field vectors due to each charge at that point have the same magnitude but opposite directions. Since electric field vectors are additive, the two vectors cancel each other out, resulting in a net electric field of zero.

    Electric Field Between Parallel Charged Plates 

    Uniform electric field between parallel charged plates

     

    Parallel charged plates produce a uniform electric field between them. The field lines between the plates are straight, parallel and equidistant from one another, indicating that the field strength is constant between the plates. Typically, the parallel plates can be charged by connecting them to a battery whose voltage creates a potential difference between the plates.

    The electric field strength in this case is given by:

    $$E = \frac{V}{d}$$​

    where

    • `V` is the potential difference between the charged plates (also the voltage of the battery) (unit: V)
    • `d` is the separation between the charged plates (unit: m)

     

    The equation above can also be used to determine the electric potential of a charge depending on its position in the electric field. Electric potential is defined as the work done on a charge per unit charge. Read more about potential here

    $$V = Ed$$

    where

    • the electric field strength `E` would be constant and independent of the charge's position 
    • `d` is the distance of a positive charge from the negatively charged plate, or of a negative charge from the positively charged plate.

    Force on Charges in Electric Fields

    When a charge is placed in an electric field, it experiences a force. The magnitude of this force `F` is given by:

    where `q` is the charge and `E` is the electric field strength.

    For a charge between charged parallel plates, this translates to:

     

    $$F = \frac{qV}{d}$$​

    The direction of force due to an electric field depends on whether the charge is positive or negative. 

     

    Force due to electric field between parallel charged plates

     

    For a positive charge, the force acts towards the negatively charged plate and parallel to the field lines (same direction as the direction of electric field). For a negative charge, the force acts towards the positively charged plate and parallel to the field lines (opposite direction as the direction of electric field).

    The acceleration experienced by a charge in an electric field would have the same direction as the force acting on it, and its magnitude can be calculated using Newton's second law:

    $$F_{\text{electric}} = ma$$

    $$a = \frac{qE}{m}$$

    Calculation Example

     

    An electron is placed midway between two parallel charged plates connected to a potential difference of 600 V and separated by 0.25 m.

    (a) Calculate the force acting on the electron due to the electric field.

    (b) What is the acceleration of the electron?

      

    Solution to part (a):

    $$F = qE$$

    $$F = (-1.602 \times 10^{-19})(2400) = -3.8 \times 10^{-16} \text{ N}$$

    $$F = 3.8 \times 10^{-16} \text{ N towards the positively charged plate}$$

    Note:

    • the value for an electron's charge is provided in the NESA HSC Physics data sheet
    • the value for force is negative because it is opposite to the direction of the electric field

     

    Solution to part (b):

    $$F_{\text{net}} = ma$$

    $$-3.8 \times 10^{-16} = (9.109 \times 10^{-31})a$$

    $$a = -4.22 \times 10^{14} \, m \, s^{-2}$$

    $$a = 4.22 \times 10^{14} \, m \, s^{-2} \text{ towards positively charged plate}$$

    Note:

    • the value for electron's mass is provided in the NESA HSC Physics data sheet
    • the direction of acceleration is the same as the direction of net force

       

    RETURN TO MODULE 4: Electricity and Magnetism