Chapter Review
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Chapter 20: Electric Potential and Electric Potential Energy Chapter Review |
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Chapter Review
In this chapter we study the potential energy associated with the electric force and the electric field. This chapter also includes our first detailed treatment of an important electrical device called a capacitor.
20-1 - 20-2 Electric Potential Energy, Electric Potential, and Energy Conservation
The electric force is a conservative force. This means that it
is useful to define a potential energy associated with this force;
we call it the electric potential energy. As always, only
changes in potential energy are measurable and so we only define
the change in potential energy, DU. The change in electric potential
energy is defined as the negative of the work done by the electric
force
DU = -WE.
Since work done depends on the force applied, and the electric
force, in turn, depends on the charge on which it is applied, the
change in potential energy must depend on the charge. Recall that
the concept of the electric field (force per unit charge) gives us
a way to handle information about the electric force without
reference to any specific charge; in a similar way we can define a
quantity, the electric potential, to handle information
about the energy without reference to a specific charge. The
electric potential difference DV is defined as
,
where q0 is a test charge that moves
from one point to the other. The SI unit of electric potential is
the J/C and is called a volt (V): 1V = J/C. Often, instead
of electric potential, the quantity V is called the voltage.
The above definition also implies that DU = q DV; a commonly used unit of energy
based on this relation is the electron-volt (eV). An eV is
the energy change experienced by an electron or proton when it
accelerates through a potential difference of 1 V
1eV = (1.60 x 10-19C)(1V) = 1.60 x 10-19J.
The above definitions imply a direct relationship between DV and the
electric field E. It works out that the electric field can be
determined from the rate at which the electric potential changes
with position,
,
where Ds represents the displacement from one position to another. The minus sign is indicative of the fact that the electric field points in the direction of decreasing electric potential. This fact means that positive charges accelerate in the direction of decreasing electric potential and negative charges accelerate in the direction of increasing electric potential. The above relation also shows that an alternative unit for electric field could be a V/m instead of N/C. In fact, both units of E are in common use, 1 N/C = 1 V/m.
The fact that we can define an electric potential energy also
means that we can extend the conservation of mechanical energy to
the motion of charged particles in electric fields. Therefore, we
can write that the sum of the kinetic and electric potential
energies must be constant Ki +
Ui = Kf +
Uf. At a given location U = qV,
therefore, the above result can also be written in terms of the
electric potential as
Ki + qVi = Kf + qVf.
Example 20.1 Motion in a Constant Electric Field: A particle of charge 22.4 mC and mass 57.2 mg is initially held at rest in a constant electric field of 133 V/m. If the particle is released and accelerates through a distance of 14.9 m, (a) determine its change in potential energy, (b) determine the change in electric potential it experiences, and (c) use energy conservation to determine its final speed.
Picture the Problem The picture shows the electric
field lines and the charged particle.
Strategy The given information allows us to calculate the work on the particle. By knowing the work we can use the definitions of electric potential energy and electric potential to do the rest.
Solution
Part (a)
| 1. Since the force and displacement are in the same direction, the work done by the electric force is: | WE = FEd = qEd |
| 2. By definition, the change in potential energy must be: | DU = -WE = -qEd = -(22.4 x 10-3C)(133 N/C)(14.9 m) = -44.4 J |
| Part (b) | |
| 1. From the definition of electric potential, we obtain DV: |  |
| Part (c) | |
| 1. According to energy conservation we must have: | |
| 2. Solving for the final speed gives: |  |
Insight For convenience I switched the unit of electric field from V/m to N/C; remember that they are equivalent.
Physlet Illustration: Electric Charge and Potential Energy |
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| In this simulation, two parallel plates, located at y= +4 mm and y= -4 mm are fixed at ± 4 Volts, respectively. The electric field lines between the plates are shown. A particle (m = 1 gm, q = +1 mC) is released from rest at a point near the top (positive) plate. Its velocity is shown in m/s. Can you verify that the total energy of the particle is conserved? | |
Hints:
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Practice Quiz
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