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Electricity Electric Potential

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Electricity
(Electric Potential)
MELCs
1. Determine the electric potential function at any point due
to highly symmetric, continuous charge distributions.
2. Solve problems involving electric charges, dipoles, forces,
fields, and flux.
3. Solve problems involving electric potential energy and
electric potentials.
Potential difference causes an electric _______.
current

To understand potential difference,

we need to understand ELECTRIC POTENTIAL.


To understand electric potential,

we need to understand WORK.


a form of energy transfer
When work is done on an object, the
energy of the object changes:

W = ΔE
W = ΔKE W = ΔPE
Electric PE
• energy a charge has due to its position in the
electric field of other charges:

EPE = Vq
EPE = electric potential energy of charge q (SI unit is J)
V = electric potential at the location of q (SI unit is J/C or V)
q = charge (SI unit is C)
Electric Potential
• at a given point, is the electric potential energy of a small
charge q situated at that point divided by the magnitude of
the charge:

V = EPE/q
• For a point charge Q,

V = kQ/r
V = electric potential at a distance r from Q
Q = source charge
r = distance from Q
At a distance r from a point charge Q, the electric potential is:
V = kQ/r
V = electric potential at a distance r from Q
Q = charge
r = distance from Q

Thus, at point A, the electric potential is VA = kQ/r.

Q r A

And a test charge at that point will have an electric


potential energy of EPE = VA q.

Q r q
If Q is positive, and if we consider another point from Q (point B), then we
have two points at different electric potentials:

Point at higher Point at lower


electric potential electric potential
Q B A

Because of the electric potential difference between A and B, the electric


field between A and B will cause a test charge q to move from B to A:

Q q A

Thus, electric potential difference causes electric charges to move.


And when charges move, there is electric current.
Sample Problem
What is the potential on the surface of
a Au nucleus? (R=6.2 fm, Z=79)

By considering the nucleus as a point charge and r=R:


V = kQ/r
= (9 x 109 N.m2/C2)(79)(1.6 x 10-19 C)/(6.2 x 10-15m)
= 183 x 105 V
= 1.8 x 107 V
The PD between the ground and a cloud in a
particular thunderstorm is 1.2 x 109 V. What is the
magnitude of the change in EPE of an electron that
moves between these points?

ΔV = ΔEPE/q
ΔEPE = ΔVq
= (1.2 x 109 V)(1.6 x10-19 C) Using the absolute value of the electron’s charge

= 1.92 x 10-10 J
= 1.9 x 10-10 J

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