Energy and Work Done on Charges in Electric Fields
This topic is part of the HSC Physics course under the section Electrostatics.
HSC Physics Syllabus
- analyse the effects of a moving charge in an electric field, in order to relate potential energy, work and equipotential lines, by applying: (ACSPH105)
Energy and Work Done in Electric Fields Explained
Work Done By Electric Field
Charges are acted on by electric forces when they are in electric fields. Electric forces acting on positive charges are directed in the same direction as electric field lines. Electric acting on negative charges are directed opposite to electric field lines.
This force results in work done as it causes displacement of a charge.
You can review the basics of work done & energy here
Work done is given by:
$$W = Fs\cos{\theta}$$
Since `F = qE`:
$$W = qEd\cos{\theta}$$
$$W = qEd$$
where
- `q` is the charge in Coulombs (C)
- `E` is the electric field strength (V m^{–1} or N C^{–1})
- `d` is the displacement of the charge due to work done on the charge (m)
- `\theta` is the angle between the displacement of the charge and the electric force. Since the force vector due to the field is always parallel to the charge's displacement, this angle is zero. (°)
Electric Potential Energy
Electric potential energy (U) is the energy that a charged particle possesses by virtue of its position in an electric field. Similar to gravitational potential energy, which depends on an object's height above the ground, electric potential energy depends on a charge's position relative to other charges or charged plates.
Electric potential energy is defined as the amount of work that can be potentially done by an electric field on a charge. The work done on the charge is equal to the change in its electric potential energy.
$$W = \Delta U$$
$$\Delta U = qEd$$
Thus, the change in a charge's potential energy can be determined by considering its displacement `d` as a result of the electric force or work done by the field.
When a positive charge is placed in an electric field, it will start to move in the same direction as electric field lines. The amount of work done on this charge during its motion is its `\Delta U`.
Vice versa, when a negative charge is placed in an electric field, it will start to move in the opposite direction as electric field lines. However, its `\Delta U` is still defined as the amount of work done by the electric field during its motion.
Specifically, when work is done on a charge by the electric field, its electric potential energy decreases and is transformed to its kinetic energy.
$$- \Delta U = \Delta K$$
What is Electric Potential?
Electric potential, `V`, is defined as the maximum work done per unit charge in an electric field:
$$V = \frac{W}{q}$$
Electric potential has a SI unit of volt (V). Volt is also equivalent to Joule per Coulomb (J C^{–1}).
Therefore, the definition of electric potential is also the change in electric potential energy of a charge per unit charge:
$$V = \frac{\Delta U}{q}$$
Electric potential is useful in determining the work done on or `\Delta U` of a charge as it moves between two points in an electric field. For a charge (`q`) moving in an electric field from a point with potential (`V_1`) to a point with potential (`V_2`), the work done `W` can also be expressed as:
$$$W = q(V_1 - V_2)$$$
$$$W = q\Delta V$$$
$Therefore, change in a charge's electric potential energy can be also expressed in terms of the potential difference between two points:$
$$$\Delta U = q\Delta V$$$
$The electric potential of a charge depends on its position, more specifically its distance from the charged plates. For a positive charge, it depends on the distance from the negatively charged plate. For a negative charge, it depends on the distance from the positively charged plate.$
Diagram shows how a positive charge's electric potential in an electric field is measured based on its distance from the negatively charged plate. For a negative charge, the distance from the positively charged plate is used to calculate its potential.
Therefore, the change in electric potential energy can be expressed in terms of the displacement over which the charge moves. This leads to the same equation we derived by considering electric force.
$$\Delta U = q(Ed_2 - Ed_1)$$
$$\Delta U = qEd$$
where `d` is the displacement of the charge due to the work done on the charge.
Work Done By Field vs Against Field
A distinction must be made between work done by the electric field on the charge and work done against the electric field:
- Work done by the field: This is work done by the electric field or force itself, which may result in an increase in the charge's kinetic energy if it moves in the direction of the force. When work is done by the field on a charge, its electric potential energy decreases, and is transformed into its kinetic energy by the law of conservation of energy.
- Work done against the field: This refers to the work that must be done to move a charge within the electric field in a direction opposite to the force exerted by the field. For example, to move a positive charge from a point of lower potential to a point of higher potential (against the direction of the field), an external force must be applied.
Conservation of Energy
Work Done By Field
In the absence of other forces, the work done by the electric field on a charge is reflected in a change in the kinetic energy of the charge. In other words, electric potential energy is transformed into the charge's kinetic energy. The mechanical energy (U + K) of the charge remains constant.
According to the conservation of energy:
$$W = -\Delta U = \Delta KE = \frac{1}{2}m(v^2 - u^2)$$
where `m` is the mass of the charge and `v` and `u` are its final and initial velocity respectively.
Work Done Against Field
When work is done by an external force against the electric field, the energy is transferred to the charge in the form of its electric potential energy.
$$W = \Delta U$$
In this instance, the charge's kinetic energy remains constant. The charge's mechanical energy (`U + K`) increases due to an increase in its `U`. While the total energy is conserved (law of conservation of energy), mechanical energy is not due to the presence of an external force.
Calculation Example 1
The diagram shows a uniform electric field produced by two parallel charged plates separated by 0.40 m. The potential difference between the plates is 2000 V. A charge (q = 3.0 × 10^{–9} C), when placed between the plates, moves from point X to Y as shown.
Calculate the work done on the charge by the electric field.
Solution:
The work done can be determined by first finding the electric potential difference between points X and Y:
$$\Delta V = Ed$$
$$\Delta V = (\frac{2000}{0.40})(0.18) = 900 \text{ V}$$
Given that, `W = q \Delta V`:
$$W = (3.0 \times 10^{-9})(900)$$
$$W = 2.7 \times 10^{-6} \text{ J}$$
Note that the work done is also the `\Delta U` of the charge; the electric potential energy of the charge decreases by `2.7 xx 10^{-6} \text{ J}`, and this amount is transformed into its kinetic energy.
Calculation Example 2
The diagram shows an electric field produced by two parallel charged plates connected to a 1200 V battery. A proton, initially at rest, is placed 4.5 cm from the positively charged plate.
(a) Explain the motion of the proton.
(b) Calculate the final velocity of the proton.
Solution to part (a):
The force due to the electric field acting on the positively charged proton is directed towards the negatively charged metal (downward in the diagram). This force causes the proton to experience rectilinear motion with uniform acceleration downward.
Solution to part (b):
As the proton moves towards the bottom plate, its electric potential energy is transformed into its kinetic energy (law of conservation of energy). In other words, work is done by the electric field on the charge.
$$\Delta U = W = q\Delta V = qEd\cos{\theta}$$
The angle between the proton's motion and the electric field direction is zero as they are parallel, so `\theta = 0`.
$$\Delta U = (+1.602 \times 10^{-19})(\frac{1200}{0.15})(0.15 - 0.045) \cos{0}$$
$$\Delta U = 1.35 \times 10^{-16} \text{ J}$$
Now consider the proton's change in kinetic energy:
$$\Delta K = \frac{1}{2}m(v^2 - u^2)$$
The proton was initially at rest, so `u = 0`.
$$1.35 \times 10^{-16} = \frac{1}{2}(1.673 \times 10^{-27})v^2$$
$$v = 4.0 \times 10^5 \text{ m/s}$$