ELECTRIC POTENTIAL ENERGY AND ELECTRIC
POTENTIAL
Pg. 573 solar car
• An electric potential energy exists that is
comparable to the gravitational potential
energy.
• Review of gravitational potential energy.
• Fig. 19.1
• The work Wab done by the gravitational force
when the ball falls from a height of ha to a
height of hb is.
• Work done by the gravitational force equals
the initial gravitational potential energy minus
the final gravitational potential energy.
• Fig. 19.2
• Test charge experiences an electric force that
is directed downward.
• As the charge moves from A to B, work is done
by this force, same way work is done by the
gravitational force.
19.2 THE ELECTRIC POTENTIAL DIFFERENCE
• F = qoE
• The work that it does as the charge moves
from A to B depends on the charge qo.
• Can be expressed per-unit-charge basis, by
dividing both sides by the charge.
• EPE/qo is the electric potential energy per unit
charge ----- called the electric potential or the
potential
• Referred by the symbol V
The electric potential V at a given point is the electric potential
energy EPE of a small test charge qo situated at the point
divided by the charge itself:
• SI unit of electric potential is a joule per
coulomb, known as a volt.
• Alessandro Volta, invented voltaic pile
(battery).
• EPE and volt are not the same.
• Electric potential energy is an energy and is
measured in joules.
• Electric potential is an energy per unit charge
and is measured in joules per coulomb, or
volts.
• We can now relate the work Wab done by the
electric force when a charge qo moves from A
to B to the potential difference Vb-Va between
the points. Combining equations 19.2 & 19.3
EXAMPLE 1: WORK, ELECTRIC POTENTIAL
ENERGY, AND ELECTRIC POTENTIAL
the work done by the electric force as the test
charge (qo = +2.0 x 10^-6C) moves from A to B
is Wab = +5.0 x 10^-5J. (a) Find the value of
the difference, ΔEPE = EPEb – EPEb, in the
electric potential energies of the charge
between these points. (b) Determine the
potential difference, ΔV = Vb – Va, between
the points.
• The positive charge in fig.
19.2 accelerates as it moves
from A to B because of the
electric repulsion from the
upper plate and the
attraction to the lower plate.
Since point A has a higher
electric potential than point
B, we conclude that a
positive charge accelerates
from a region of higher
electric potential toward a
region of lower electric
potential
• A negative charge placed between the plates
behave in the opposite fashion.
• A negative charge accelerates from a region of
lower potential toward a region of higher
potential.
• Car Battery
• + (a)higher potential than – (b) Va – Vb = 12V
EXAMPLE 3: OPERATING A HEADLIGHT
The wattage of the headlight on the last slide is
60W. Determine the number of particles,
each carrying a charge of 1.60 x 10^-19 C (the
magnitude of the charge on an electron), that
pass between the terminals of the 12 V car
battery when the headlight burns for one
hour.
• The number of particles is the total charge
that passes between the battery terminals in
one hour divided by the magnitude of the
charge on each particle. The total charge is
that needed to convey the energy used by the
headlight in one hour. This energy is related
to the wattage of the headlight, which
specifies the power or rate at which energy is
used, and the time the light is on.
WHAT YOU KNOW (about this problem)
Volt is used for measuring electric potential
difference.
Volt can be used to describe an electron or a
proton, electron volt (eV).
One electron volt is the magnitude of the
amount by which the potential energy of an
electron changes when the electron moves
through a potential difference of one volt.
• One million (10^+6) electrons volts of energy
is referred to as one MeV, and one billion
(10^+9) electron volts of energy is one GeV,
where the G stands for the prefix “giga.”
EXAMPLE 4: THE CONSERVATION OF ENERGY
A particle has a mass of 1.8 x 10^-5 kg and a
charge of +3.0 x 10^-5C. It is released from
rest at point A and accelerates until it reaches
point B. The particle moves on a horizontal
straight line and does not rotate. The only
forces acting on the particle are the
gravitational force and an electrostatic force.
The electric potential at A is 25V greater than
that at B; in other words Va – Vb = 25V.
What is the translational speed of the particle at
point B?