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Electrical Energy and Potential
Difference
When two charges interact, there is a force
between them. This force creates potential
energy.
This electric PE is a form of mechanical
energy, like KE, GPE and PEelastic
Charges moving in a uniform electric field
+
-
+
-
A + charge moving against the electric field gains PE
A + charge moving with the electric field loses PE
A - charge moving against the electric field loses PE
A - charge moving with the electric field gains PE
DPEelectric = -qEDd
Measured in Joules
Dd is in the direction of the electric field
Electrical Potential Energy for a pair of
charges
q1q2
PEelectric  kC
r
Electric Potential is the amount of
PEelectric divided by the amount of
charge
PEelectric
V
q
Potential Difference is the change in V
DPEelectric
DV 
q
Potential Difference is measured in
Volts (V), which equals 1 J/C
It can also be expressed as DV = -EDd
or DV = kCq/r or DV = W/q
Sample Problems
A uniform electric field is directed in the + x direction at
250 N/C. A 12mC charge moves from the origin to the point
(20.0 cm, 50.0 cm). What is the change in PE?
Given
DPE
= -qEDd
q = 12 x 10-6 C
moving with
field, so will
lose PE
E = 250 N/C
d = .200 m
electric
= (-12 x 10-6 C)(250 N/C)(.200 m)
= -6.0 x 10-4 J
A proton is released from rest in a uniform electric field
of 8.0 x 104 V/m and moves .50 m. Find the potential
difference and DPE for these positions.
Given
a) DV = -EDd
E = 8.0 x 104 V/m
DV = -(8.0 x 104 V/m)(.5 m)
d = .5 m
DV = -4.0 x 104 V
b) DPE = DVq
q = 1.60 x 10-19 C
DPE = (-4.0 x 104 V)(1.60 x 10-19 C)
DPE = -6.4 x 10-15 J