Download Electric potential

University of Puerto Rico at Humacao
FISI3012 – Professor J. C. Cersosimo
Examples: Electric potential
Equipotential line or equipotential surface
Observe that if a charged particle moves perpendicular to electric field
lines, no work is done on it, and its electric potential energy does not
change. Along such a path the electric potential energy and the electric
potential, V, remain constant.
Two charge: –q and +q. The equipotentials are perpendicular to electric
field lines. V1  V 2 , because equipotential surface 1 is closer to the
positive charge than is furface 2.
Two plates are separated a distance d. The actual equipotentials surfaces
within the parallel plates are planes parallel to those plates. The zero
potential is choose at the negative plate.
Problem 1:
Find the potential due to an electric dipole far from the dipole.
Solution:
1 1
r r 
V  kq     kq  1 2 
 r1 r2 
 r1r2 
r2  r1  2a cos 
r1r2  r 2
V  kq
2a cos p cos

r2
4 0r 2
p is the dipole moment, p  2aq
1
University of Puerto Rico at Humacao
FISI3012 – Professor J. C. Cersosimo
2
Problem 2
For the charged metal plates of figure , suppose that the plate separation D is 5.0 cm, the
distance d from A to B is 3.5 cm, and the electric field is 1.4×103 N/C.
a) How much work must I do to move a charge of +1.20 μC from B to A?
b) What is the differential difference VAB between A and B?
c) What is the differential difference between the two metal plates?
d) How much work must I do to move a charge from A to B?
Solution:
a) The work I must do to move the charge from B to A is:
 
 
W  F  d  qE  d
Since F and d are in the same direction
W  1.2 106 C 1.4 103 N / C 3.5 102 m  5.88 105 J
The work goes into increased PE of the charge q at its new position (A) in the field.
b) The potential difference between A and B is
VAB  VA  VB 
W 5.88  10 5 J

 49.0 J / C  49.0V
q 1.2  10 6 C
A is a higher potential than B, since the electric field is directed from A to B.
c) Since the electric field between the plates is constant, the potential difference is
directly proportional to distance. Therefore
VAB / d  V / D , where V is the potential difference between the plates and D is
the distance between the plates. Therefore
V
D
5.0cm
49.0Volt   70.0V
VAB 
d
3.5cm
d) Since the electrostatic field is conservative, the work done in taking a charge
around a closed path must be zero. Therefore the work in taking the charge from A to
B must be the negative of the work required to move the charge from B to A, as
5
found in part (a). Therefore W  5.88 10 J
University of Puerto Rico at Humacao
FISI3012 – Professor J. C. Cersosimo
3
Problema 3
An electron in a TV set is accelerated toward the screen by a potential difference of 1000
V. The screen is 35 mm from the electron source.
a) How much work is done by the electric field?
b) What is the speed of the electron when it strikes the screen?
Solution:
a) Since the work done per unit charge by the field is,
W
V ;
q
W  qV  1.5 10 19 C 1000V   1.6 1016 J
Another approach is to note that an electron moving through a potential difference of
1000V has energy of 1000 eV.
b) Since no external force act, by the principle of conservation of energy all this
work must be converted into kinetic energy of the electron:
KE  12 mv2  1.6 10 16 J
And
 21.6  10 16 J 
v

31
 9.11  10 kg 
1/ 2
 1.87  10 7 m / s
Note that the distance between source and screen is of no importance here. The
crucial factor is the potential difference between sources and screen; this completely
determines the work done on the electron, and hence its final KE and speed.
University of Puerto Rico at Humacao
FISI3012 – Professor J. C. Cersosimo
4
Problem 4
A neutral water molecule (H2O) is composed of ten electrons, ten protons and eight
neutrons, separated by a distance of 3.910-12 m.
a) What is the magnitude of their electric dipole moment?
b) What is the maximum torque exerted on the dipole when it is placed in a uniform
electric field of magnitude 3.0105 V/m?
c) What is the potential energy when it is oriented 30o to the electric field?
d) How much work does the field when the dipole rotates from =120o to =30o?
Solution:
Part a)
The magnitude of the electric dipole momentum is
p  qd  1.6  10 19C 3.9  10 12 m   6.24  10 31C  m
Part b)
The maximum torque is exerted on the dipole when its dipole moment is
perpendicular to the applied electric field, that is, when =90o
 max  pE sin 90 o  6.24  10 31C  m 3  10 5 N / C   18.72  10 25 N  m
Part c)
 
From equation: U   p  E , the potential energy of
the dipole at 30o is
U 30   pE cos30 o
 6.24  10 31C  m 3  10 5 N / C 0.866
 16.21  10 26 J
Part d)
The work done by the field when the dipole rotates between two orientations is the
negative of the change in the dipoles potential energy:
W  (U 30  U 120 )  16.21  10 26 J    8.11  10 26 J   24.32  10 26 J