Download Chapter 29. The Electric Potential

Chapter 29. The Electric Potential
At any time, millions of light
bulbs are transforming
electric energy into light and
thermal energy. Just as
electric fields allowed us to
understand electric forces,
Electric Potential allows us
to understand electric
energy.
Chapter Goal: To calculate
and use the electric potential
and electric potential energy.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
What are the units of potential difference?
A.  Amperes
B.  Potentiometers
C.  Farads
D.  Volts
E.  Henrys
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
What are the units of potential difference?
A.  Amperes
B.  Potentiometers
C.  Farads
D.  Volts
E.  Henrys
The work W to move charge
q through a potential
difference V is:
W= qV
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
New units of the electric field were
introduced in this chapter. They
are:
A.  V/C.
B.  N/C.
C.  V/m.
D.  J/m2.
E.  Ω/m.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
New units of the electric field were
introduced in this chapter. They
are:
A.  V/C.
B.  N/C.
C.  V/m.
D.  J/m2.
E.  Ω/m.
The (differential) work to
move charge q through a
potential difference dV is
dW= q dV = - q E dx
E=- dV/dx
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The electric potential inside a capacitor
A.  is constant.
B.  increases linearly from the
negative to the positive plate.
C.  decreases linearly from the
negative to the positive plate.
D.  decreases inversely with distance
from the negative plate.
E.  decreases inversely with the
square of the distance from the
negative plate.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The electric potential inside a capacitor
A.  is constant.
B.  increases linearly from the
negative to the positive plate.
C.  decreases linearly from the
negative to the positive plate.
D.  decreases inversely with distance
from the negative plate.
E.  decreases inversely with the
square of the distance from the
negative plate.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Differential work to
move charge q
through a potential
difference dV is
dW= q dV = -q E
dx
-dV/dx = E
If E is constant, V
V=V0-Ex
We define the electric potential V (or, for brevity, just the
potential) as
Charge q is used as a probe to determine the electric
potential, but the value of V is independent of q. The
electric potential, like the electric field, is a property of
the source charges.
The unit of electric potential is the joule per coulomb,
which is called the volt V:
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The electric potential inside a parallel-plate capacitor is
where s is the distance from the negative electrode.
The electric potential, like the electric field, exists at all
points inside the capacitor.
The electric potential is created by the source charges on
the capacitor plates and exists whether or not charge q is
inside the capacitor.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
QUESTIONS:
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Note: Dropping a proton through the 500 Volt potential
difference will give it a high speed!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Consider two point charges, q1 and q2, separated by a
distance r. The electric potential energy is
This is explicitly the energy of the system, not the energy of
just q1 or q2.
Note that the potential energy of two charged particles
approaches zero as r approaches infinity .
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Let q be the source charge, and let a second charge q', a
distance r away, probe the electric potential of q. The
potential energy of the two point charges is
By definition, the electric potential of charge q is
The potential extends through all of space, showing the
influence of charge q, but it weakens with distance as 1/r.
This expression for V assumes that we have chosen V = 0 to
be at r = infinity.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
QUESTIONS:
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Consider two point charges, +e and -e, separated by a
distance r = 1.0e-10 m. The electric potential energy is
Define unit eV = electronvolt = 1.602 e-19 J, the energy a
charge +e acquires in falling through a potential difference
of 1 volt. The scale of potential differences in atoms is volts
and the energy scale is eV. THAT is why atoms radiate/
absorb visible light quanta of similar energy.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The electric potential V at a point in space is the sum of the
potentials due to each charge:
where ri is the distance from charge qi to the point in space
where the potential is being calculated.
In other words, the electric potential, like the electric
field, obeys the principle of superposition.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
2H
H2
To remove an electron from a neutral hydrogen takes 13.6
eV. Combining two hydrogen atoms permits both electrons
to be between and close to two protons => binding energy
of order eV released per atom pair. Binding of unlike atoms
is asymmetric but the energy scale is the same: eV/bond.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Two grams of H combine with 1/16 gm of O to form water.
If 2H+O=>H2O released E=1 eV of “chemical energy,”
what is the total energy release in joules and how high
would that lift one kg subject to earth gravity?
Note: Demonstration ignited balloon filled with a mixture of H2
and O2.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Two grams of H combine with 1/16 gm of O to form water.
If 2H+O=>H2O released E=1 eV of “chemical energy,”
what is the total energy release in joules and how high
would that lift one kg subject to earth gravity?
The number of molecules formed = the number of protons in
one gm = the inverse of the proton mass (1.67e-24 gm) =
Avogadro’s number N=6e23. The energy released is
Q=N*E=6e23*1 eV
That could raise 1 kg by Q=mgh=>h=Q/mg = 1000 m. Sweet!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
How much kinetic energy in eV is required to throw a
proton at a Pb (Z=82) nucleus so the proton comes to rest
just at the surface (r= 6 fm)?
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
How much kinetic energy in eV is required to throw a
proton at a Pb (Z=82) nucleus so the proton comes to rest
just at the surface (R= 6 fm)?
The nucleus has charge +Ze. To go from r=infinity to r=R,
converting KE entirely to PE at r=R requires
To achieve this KE, the proton must be dropped through a
potential difference of 20 MV=>Van de Graff accelerator.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.