Work and Electric Potential
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Transcript 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