Download Chapter 25 Review: Electric Potential

Chapter 25 Review:
Electric Potential
Matthew Volk
Background / Summary
Electric potential is the potential energy per unit charge of a charged object in an electric field. This chapter covers
potential energy, electric potential difference (voltage), voltages in constant electric fields, voltages due to point
charges, voltages due to continuous charges, and the relationship between voltage and electric field strength. For a
quick summary, look at the formulae table, but for a more in-depth description of each equation look at the topic
summaries.
Potential Energy and Voltage
 The change in potential energy of a charged object moving
through an electric field is given by the equation:
Formulae to Know:




Voltage is defined as the change in potential energy per unit
charge:

Work is given by the following equation:



Voltages in Constant Electric Fields
 For constant electric fields, E can be removed from the integral:



Protons in Electric Fields
High V
High U
Low V
Low U

Field lines – Field lines always go from high to low voltage.
Positive charges follow them, going from high U and high V to
low U and low V. Negative charges move in the opposite
direction as them, going from high U to low U, but low V to
high V (remember is negative).
You can use these equations to find velocity of a particle in an
electric field using conservation of energy:
Electrons in Electric Fields
High V
Low U
Low V
High U
Voltage due to Point Charges
 Voltage due to a point charge is given by the following
relationship. For multiple point charges, you sum up the
resulting from each charge.
’s
Millikan Oil Drop Experiment
qE
mg

Potential energy of a system of particles is given by similar
equations. For two particles, the first equation applies. For 3 or
more, the equation needs to be expanded as shown:
Definitions to Know:



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Voltage (Electric Potential) – defined
as the potential energy per unit charge
of an object in an electric field
Electric Field Lines – Always point in
the direction of decreasing voltage
Equipotential Surfaces – surfaces
over which all points have the same
electric potential
Millikan Oil Drop Experiment –
Experiment that proved charge is
quantized and calculated the elemental
charge unit
Tip: For electric potential due to various
common charge distributions, like spheres
or charged rings, look at table 25.1 (page
786) in the book. It gives a comprehensive
list of the results of the equations given
here that might be useful to memorize if
you prefer the memorization approach. I
recommend memorizing the sphere ones at
the least, because it’s useful to not have to
derive them each time.
Electric Field and Voltage
 The magnitude of the electric field is equal to the opposite of
the change in voltage over a set distance. In other words,

For objects in 2D or 3D space, you have to take partial
derivatives, or you can use the Del operator for shorthand:
Voltage due to a Continuous Charge Distribution
 To find the voltage change due to a continuous charge
distribution, you take the equation used for multiple charges
and change the sigma to an integrand:
Miscellaneous Stuff to Know
 The voltage is constant across the surface of a charged
conductor. In other words,

The Millikan Oil Drop Experiment – This experiment is famous
because it proved that charge is quantized and measured the
charge of an electron. Essentially, they sprayed ionized
particles of oil into a variable electric field controlled by a dial.
They would change the field strength until some particles
remained suspended or stopped accelerating, in which case
they knew that the force of gravity was equal to the force of the
electric field:
From there, they calculated q on many different particles,
found the smallest q to equal qe, and plotted them to find a
stepwise function, proving that charge is quantized.
Problems
1.
[Easy] Three point charges are located along the x-axis as follows: a +2.00 μC point charge is located at + 3.00 mm, a 0.50 μC
point charge is located at – 5.00 mm, and a +5.00 μC point charge is located at the origin. a) What is the potential energy of this
system of point charges? b) What is the resultant voltage at x=1.00 cm due to this system of point charges?
2.
[Medium] A particle enters an electric field with an initial velocity to the left of 1.50 x 10 6 m/s. The electric field lines point to
the right. Assume there is no loss of energy. What voltage drop must this particle travel through before its velocity reaches 0 if it
is a proton? If it is an electron?
C
B
[Hard] Two spherical conducting shells are arranged as shown in the figure to the right. The inner radius
3.
r1 = 5.00 cm and has charge 4.00 μC. The outer shell has radius r2 = 10.0 cm and has charge -6.00 μC.
a) What is the electric field in regions A, B, and C? b) What is the electric potential V in regions A, B,
and C?
A
r1
r2
Problem 1
q1 = +0.50 μC
q2 = +5.00 μC
q3 = +2.00 μC
a potential energy, I use the equation for potential energy of a system of particles:
a) To find
b) To find voltage, I use the voltage equation:
Problem 2
a) This is an energy problem. So, start with the conservation of energy equation for electric field problems:
b) Because an electron will accelerate in the direction opposite that of the field lines, it will speed up through this
field, not slow down. Therefore, no change in voltage will cause this electron to stop in this field. (Not possible.)
Problem 3
a) In area A, we know from Gauss’s Law that, because there is no charge internal, there is no electric field. So,
C
B
In area B, we can apply the equation for the electric field outside a sphere (see ch. 24
if you need help with Gauss’s Law)
A
r1
r2
In area C, we use the same equation, noting that the total enclosed charge is now the sum of the two charges.
b) In area C, we can use the following equation to get a function of V(r):
VB equals VC at r2 plus the effects of the voltage in area B as well. In other words, we simply add the voltage at r2 to the
integral of the electric field equation derived in part a:
Finding VA is significantly easier. The voltage inside area A is the same as the voltage at points a distance r1 from the
center. Because r1 is within the domain of the equation we just created for VB, we can find VA by taking
. This
makes sense because we know that the voltage inside area A should be constant.