1. Work and Electric Potential
Physics Chapter 22 B

2. Work Remember: For an electric charge: Vb – Va = W/q
Work is using a force to move something a distance
W = Fd
For an electric charge:
Work is done when a force moves a test charge farther away from a negative source charge
We are increasing the potential energy of the test charge
Work is proportional to the magnitude of the charge moved
Because electric force is related to the magnitude of the charges involved!
Vb – Va = W/q

3. Electric Potential
Difference in electric potential measures the effect of location within the electric field
Low potential:
High potential:
Only changes in electric potential are related to work done!
Electric potential is measured in Volts
1 volt = 1 Joule/Coulomb + + + +

4. Uniform Electric Fields
Uniform fields can be created using parallel flat plates with charges
Work done to move a charge in this field is Fd
V = Fd/q (potential is work/charge)
Since F/q equals E (field intensity):
V = Ed

5. Problem p 372
Two parallel plates are m apart. The electric field intensity between them is 6.00 x 103 N/C. What is the potential difference between the plates?
What we know:
D = m E = 6.00 x 103 N/C
Equation:
V = Ed
Substitute:
V = (6.00 x 103 N/C)(0.500 m)
Math:
V = 3.00 x 103 Nm/C = 3.00 x 103 J/C

6. Problem p 372 Continued
Two parallel plates are m apart. The electric field intensity between them is 6.00 x 103 N/C. What work is done moving a charge equal to that on one electron from one plate to the other?
What we know:
Q = 1.6 x C V = 3.00 x 103 V
Equation:
V = W/q so W = Vq
Substitute:
W = (3.00 x 103V)(1.6 x C)
Math:
W = 4.8 x CV = 4.8 x J

7. Problem p 373
A voltmeter measures the potential difference between two parallel plates to be 60.0 V. The plates are m apart. What is the magnitude of the electric field intensity?
What we know:
V = 60.0 V d = m
Equation:
V = Ed so E = V/d
Substitute:
E = (60.0 V)/(0.030 m)
Math:
E = 2.0 x 103 V/m = 2.0 x 103 N/C