1. Electric Potential Energy or Potential Difference (Voltage) Recall the idea of Gravitational Potential Energy: lifting an object against gravity requires work, and energy is transferred to the object. The same idea can be used to explain electric potential.

2. Consider two charged objects found between two charged plates, each experiences a force to the right of magnitude: F = qE

3. Object 1: Suppose it is free to move through a distance d, from B to A. Since a certain force acts on an object through a distance d, a certain amount of work is done on the charged object. W = F e d Object 2: We could force this charge to the left if we grabbed it and exerted a force on it throughout the distance d. This would mean we would have to do work on the charge. (W = F e d) * If work is done on a charge when it freely moves or is forced to move from one point to another, these two points are said to differ in their electric potential, The amount of work done is a measure of this difference in electric potential or electric potential energy.

4. Potential Difference (V) The potential difference between two points in an electric field is the amount of work done per unit of charge as a charge is moved between two points. Mathematically: Potential Difference = Work Done Charge Rearrange toW = qV

5. The volt – one volt is the potential difference between two points in an electric field when one joule of work is done in moving a charge of 1 C between these points.

6. Relationship between Potential Difference (Voltage) and Electric Field Strength We have seen that: Remember 1 J = 1 Nm So 1 V = 1 Nm/C This gives us that 1 V/m = 1 N/C

7. so Volts / metre – Newtons / Coulomb = units of electric field strength (E) Thus, the electric field strength which we previously expressed in units of N/C can also be expressed in units of volts / metre. This leads us to the formula r = distance (m)

8. When E is expressed in volts / metre it gives the charge or difference in electric potential energy per unit of distance, and is called the potential gradient of the electric field.

9. Sample Problem A charged particle of mass 4.0 x 10 -15 kg is balanced between charged plates. The battery is rated at 5000 V and the plates are 4.0 cm apart.

10. Find electric Field Strength electric force on the particle amount of charge on the object number of elementary charges on the particle solution