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Electric Fields, Dipoles and Torque
Challenge Problem Solutions
Problem 1:
Three charges equal to –Q , +Q and +Q are located a distance a apart along the x axis (see sketch). The point P is located on the positive y -axis a distance a from the origin.
(a) What is the electric field E
G
at point P?
(b) (f) (enter one letter) is the correct field line representation for this problem
Problem 1 Solution:
(a) If we number the charges from left to right, then at point P , the E field due to the charges are
( )
( ) ( )
( )
( )
( ) ( )
(^1 3 )
(^2 )
(^3 3 )
e e
e
e e
Q Q
k a a k a a
Q k a Q Q k a a k a a
E i j i j
E j
E i j i j
G
G
G
Thus, the total electric field is 1 2 3 2 ˆ^2 ˆ 2 e e
Q Q
k k a a
E = E + E + E = − i + j
G G G G
(f) is the correct field line representation for this problem You can tell this because the center and leftmost charges are of opposite signs (field lines start on one and end on the other), while the center and rightmost charges are of the same sign (their field lines repel each other).
9 2 2 6 6 3 2 (^32 ) 0
1 | | 8.99^10 N m^ C^ 7.00^10 C^ 4.00^10 C 1.01 N (^4) 0.500 m
q q F
πε r
× ⋅ × −^ × −
In Cartesian coordinates, the forces can be written as
( ) (^ )^ (^ )
31 31
32 32
cos 60 ˆ^ sin 60 ˆ^ 0.25 N ˆ^ 0.44 N ˆ
cos( 60 )ˆ sin( 60 )ˆ 0.51N ˆ 0.87 N
F
F
= ⎡^ − ° + − ° ⎤= −
F i j i
F i ) ˆ
j
j i j
G
G
The total force is
F = F 31 + F 32 = ( 0.76 N ) ˆ i^ −( 0.44 N)ˆ j
G G G
The magnitude of F is
G
2 2 (0.76 N) 2 ( 0.44 N) 2 0.87 N
F = Fx + Fy = + − =
and the angle with respect to the + x axis is
tan 1 tan 1 0.44 N 30 0.76 N
y x
F
F
ϕ −^ −
⎝ ⎠ ⎝^ ⎠
Problem 3:
Cesium Chloride is a salt with a crystal structure in which cubes of Cs+^ ions (side length a ~ 0.4 nm) surround Cl-^ ions, as pictured at right.
(a) What is the magnitude of the net electrostatic force on the Cl-^ ion due to its eight nearest neighbor Cs+^ ions?
(b) Occasionally defects arise in which one of the Cs+^ ions is missing. We call this a vacancy. In this case, what is the magnitude and direction (relative to the vacancy) of the net electrostatic force on the Cl-^ ion due to its remaining seven
Problem 3 Solution:
(a) By symmetry the electrostatic force from each Cs+^ ion is cancelled by its partner opposite the Cl-^ ion, so the net force is zero.
(b) Taking away a Cs+^ ion to create a vacancy is, electronically, the same thing as adding a negative charge to the system at the location of the vacancy (this will cancel out the positive charge of the ion, essentially removing it). This negative charge, q = - e , will
create a repulsive force on the Cl-^ ion, a distance (^) ( ) 2 d = 3 a 2 away, of magnitude:
( )
( ) ( )
( )
2 2 9 2 -2^192 9 2 2 10 2
9 10 Nm C 1.6 10 C 1.9 10 N (^3 2) 3 2 10 m
ke ke F d (^) a
− − −
× ×
= = = = ×
×
That is pretty small. We probably aren’t justified in ignoring gravity in this case. But put on a million electrons so we are up near a gram and this becomes reasonable.
Problem 5: Two massless point charges +9 Q and - Q are fixed on the x-axis at x = - d and x = d:
(a) There is one point on the x-axis, x = x 0 , where the electric field is zero. What is x 0?
(b) A third point charge q of mass m is free to move along the x-axis. What force does it feel if it is placed at x = x 0 (the location you just found)?
(c) Now q is displaced along the x-axis by a small distance a to the right. What sign of charge should q be so that it feels a force pulling it back to x = x 0?
(d) Show that if a is small compared to d ( a << d ) q will undergo simple harmonic motion. Determine the period of that motion. [NOTE: The motion of an object is simple harmonic if its acceleration is proportional to its position, but oppositely directed to the displacement from equilibrium. Mathematically, the equation of
motion can be written as d x^2 / dt^2 = − ω 2 x , where^ ω^ is the angular frequency. See
Review Module E for more detail. HINT: a / d is REALLY SMALL. Taylor expand]
(e) How fast will the charge q be moving when it is at the midpoint of its periodic motion?
Problem 5 Solution:
Two point charges 9 Q and - Q are fixed on the x-axis at x = - d and x = d respectively.
(a) The charges are of opposite sign, so the field zero is not between them but to one side or the other. Since the 9 Q charge at x = -d is bigger, the zero will be to the right of the - Q charge ( x 0 > d). Now we can solve:
( 0 ) 2 2 2 0 0 0 0 0 2 2 0 0 0 0 0
Q Q Q Q
x x d x d x d x d
x d x d x d x d x d
πε
⎝ +^ −^ ⎠ +^ −
E
G
2
E
G G
(b) Since F = q and E ( x 0 (^) )= 0 , the charge feels no force at x = x
G
9 Q - Q
d
y
x
d
2 max (^3) (^30)
q Q v a a md
which is the speed of q at the midpoint of its motion (x = d ).
Problem 6:
Consider two equal but opposite charges, both mass m , on the x-axis, + Q at ( a ,0) and – Q at (- a ,0). They are connected by a rigid, massless, insulating rod whose center is fixed to
a frictionless pivot at the origin. This is a dipole. A uniform field E = E ˆ i
G
is now applied.
(a) What is the force on the dipole due to this external field?
Now the dipole is rotated and held at a small angle θ 0 (c.c.w.) from the x-axis.
(b) Now what is the force on the dipole?
(c) How much did the potential energy of the dipole change when it was rotated?
(d) What is the torque on the dipole?
(e) Now the dipole is released and allowed to rotate due to this torque. Describe the motion that it undergoes (i.e. what is its angle θ( t )?)
(f) Where is the positive charge when it is moving the fastest? How fast is it moving?
Problem 6 Solution:
(a) A dipole in a uniform field feels no force (the force on the positive charge is equal and opposite to the negative charge).
(b) The field is still uniform so the force is still zero.
(c) The change in potential energy of a charge q is given by: B
A
Δ U = q Δ V = − q ∫ E ⋅ d s
G G
For example, our positive charge changes its potential energy by:
0 0
0
0 0
0 0
ˆ ˆ ˆ (^) sin ˆ (^) cos ˆ sin
cos 1 cos
U Q E ad QEa d QEa d
QEa QEa
θ θ θ
θ θ θ θ
0
0
= = =
∫ i^ θ^ ∫ i^ i^ j ∫^ θ
Similarly, the negative charge moves from θ = π to θ = π + θ 0 , changing its energy by:
Δ U = − QEa ( cos π − cos ( π − θ 0 )) = − QEa ( − + 1 cos ( θ 0 )) = QEa ( 1 −cosθ 0 )
So both charges change their potential energy by the same, and the total change in the
dipole’s potential energy is: Δ U dipole = 2 QEa ( 1 −cos θ 0 )
Note that we could write this as follows:
Δ U dipole = − 2 QEa ( cos θ 0 − cos 0) = − pE ( cos θ 0 − cos 0)= Ufinal − Uinit
We also could have done this equating potential energy and kinetic energy (and realizing that both masses must be moving at the same speed):
( ) ( ) 1 2 1 2 0 2 0 2
0
U (^) dipole 2 QEa 1 cos 2 QEa Kinetic Energy 2 m
QEa v m
v
where again we had to use a small angle approximation that cos θ 0 ≈ 1 − 12 θ 02.
MIT OpenCourseWare http://ocw.mit.edu
8.02SC Physics II: Electricity and Magnetism
Fall 2010
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