Show that the electric field intensity at any point in the electric field is equal to the rate of change of potential difference with respect to distance.
Answers
Answer:
The electric field is a measure of force per unit charge; the electric potential is a measure of energy per unit charge.
For a uniform field, the relationship between electric field (E), potential difference between points A and B (Δ), and distance between points A and B (d) is:
E
=
−
Δ
ϕ
d
If the field is not uniform, calculus is required to solve.
Potential is a property of the field that describes the action of the field upon an object.
Key Terms
electric field: A region of space around a charged particle, or between two voltages; it exerts a force on charged objects in its vicinity.
- electric potential: The potential energy per unit charge at a point in a static electric field; voltage.
Explanation:
- mark as brain least
Electric field intensity is the measure of the electric field at any point.
The potential difference between the two points is the amount of energy acquired by a unit charge when it is moved from one point to another
Work done = F×d
Potential difference = energy/charge
From Columbus law, F = qE
Substituting the value of work done in the potential difference equation
V₂ - V₁ = F×d/q
V₂-V₁ is the potential difference between two points.
V₂ - V₁= q×E×d/q
V₂ - V₁ = Ed
Therefore, E = V/d, that is, Electric field intensity is equal to the rate of change of potential difference with respect to distance.