Download Electric Potential

Lesson 39

Consider an electric field between two equally
but oppositely charged plates a and b whose
separation is small. The electric field would
be uniform between these plates.
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a
b
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a
+q
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If we release a charge +q from the positive
plate b the electric force would accelerate the
charge toward the negative plate.
b
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
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The electric field will do work to move the charge
across the electric field.
◦ W = ΔKE
The charge has high Electrical Potential
Energy near b
As the charge move closer to a, the Electrical
PE decreases while the KE increases
◦ The size of PE is dependent on the size of the
electric field

Electric Potential – the electric potential per
unit charge between the plates
◦ V for voltage (the symbol and unit are the same)

V = PEelectric/q
◦ It is impossible to find the V at any point in an
electric field

We can only measure the difference in
potential between two points.
 Vba
= Va - Vb
 Vba = PEa/q – PEb/q
 Vba = ΔPE/q
 ΔPE = qVb
 ΔU = qV
 Where ΔU is the transformation
of PE to KE and vice versa.

Vba = W/q
◦ where W is the net work to move the charge across
the magnetic field.
◦ The unit for voltage is Joules/Coulomb, or simply
Volts in honor of Alessandro Volta.

We define the direction of the electric field as
originating at areas of high potential (the
positive plate in the picture) and ending at
areas of low potential (the negative plate)

Often the areas of low potential are assigned
a Zero value.
◦ Called ground.

A positive charge is moved from the positive
plate to the negative plate
◦ Does the charge gain potential (+voltage) or lose it
(-voltage)
◦ Is the change in the PEelectric positive or negative?
◦ Is the work done by the electric field positive or
negative?



Negative
Negative
Positive - Wnet is positive (force is in the
same direction as displacement)

A positive charge is moved from the negative
plate to the positive plate.
◦ Does the charge gain potential (+voltage) or lose it
(-voltage)
◦ Is the change in the PEelectric positive or negative?
◦ Is the work done by the electric field positive or
negative?



Positive
Positive
Negative(force is in opposite direction of
displacement)

A negative charge moves from the negative
plate to the positive plate.
◦ Does the charge gain potential (+voltage) or lose it
(-voltage)
◦ Is the change in the PEelectric positive or negative?
◦ Is the work done by the electric field positive or
negative?



Positive
Negative
Positive

A charge of –6.0μC is moved from ground to
a point where the potential is +75V. a) What
is the change in PE? b) Is the work done by
the electric field positive or negative? c) How
much work is done by the electric field?

-4.5x10-4J, Positive, 4.5x10-4J

An electron, released from plate A accelerates
to plate B acquiring 3.2x10-15J of KE. a) What
is the potential difference across the plates?
b) Which plate has the higher potential?

KE = -PE, 2.0x104V, B is at a higher potential


E = V/ Δd – for uniform electric fields! (E can
be expressed as N/C or V/m)
Does not apply to fields caused by point
charges!

The joule is a large quantity of energy when
dealing with protons and electrons. The
electron volt (eV) is much smaller. It often
provides a more meaningful magnitude for
expressing energies

1eV is defined as the amount of energy
required to move an electron through a
potential difference of 1Volt
◦ An electron that accelerates through a potential
difference of 100V will lose 100eV of electric
potential energy and gain 100eV of kinetic energy.
The eV is not a proper SI unit so we must convert to
Joules when performing calculations involving SI
units.
◦ 1.0eV = 1.6x10-19J

Two oppositely charged plates A and B are
0.010m apart. The E between the plates is
4000N/C directed from plate B to A. A
proton is released from rest at the high
potential plate and it accelerates to the low
potential plate. a) Determine the voltage
difference between the plates. b) Which plate
has a higher potential? c) Calculate the
kinetic energy, in eV, of the proton when it
reaches the low-potential plate.

V = 40V, Plate B is high potential, KE = 40eV
NO

For point charges, the electric potential
infinitely away from a point charge is zero.
◦ V = kQ/r or V = 1/4πεoQ/r


The electric potential near a negative charge
is negative (less than zero).
The electric potential near a positive charge is
large and decreases to zero at large distances

A +30μC QL is placed 30cm from an identical
A +30μC charge QR. a) Determine the
magnitude of the potential at point P, which
lies in a line between the two charges and is
10cm from QL. b) Determine the electric field
at point P. c) Determine the work done by
the electric field in moving a +0.50 μC charge
from point P to point M(the midpoint).

4.0x106V, 2.0x107N/C to the right, -0.22J.