Download Work and Electric Potential

Work and Electric Potential
APPLIED PHYSICS AND CHEMISTRY
ELECTRICITY LECTURE 4
Work
 Remember:
 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
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
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
Problem
 Two parallel plates are 0.500 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 = 0.500 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
Problem Continued
 Two parallel plates are 0.500 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 10-19 C
 Equation:

V = W/q so W = Vq
 Substitute:

W = (3.00 x 103V)(1.6 x 10-19 C)
 Math:

W = 4.8 x 10-16 CV = 4.8 x 10-16 J
V = 3.00 x 103 V
Another Problem
 A voltmeter measures the potential difference between
two parallel plates to be 60.0 V. The plates are 0.030 m
apart. What is the magnitude of the electric field
intensity?
 What we know:

V = 60.0 V
d = 0.030 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