Discussion Question 2A
P212, Week 2
Electric Field Lines
Electric fields are force fields produced by electric charges. They can extend throughout space, and
influence other charges (and currents and magnetic fields, as you will discover later in the course).
Electric Field Lines graphically represent electric fields. They show you the direction and relative
magnitude of force that a test charge would feel in the field region.
(a) The configuration at right shows some points
around a positive charge Q. If a positive test charge q
is placed at point A, at a distance rA from Q, calculate
the magnitude of the force on the test charge in terms
the given variables.
D
B
Q
C
A
|FA| = kqQ/rA2
Draw the direction of the force with an arrow on the diagram.
(b) Would the magnitude of the force be bigger or smaller if the charge was placed at points B,C, or D?
Force would be bigger at B, smaller at C or D
Draw the direction of these forces on the diagram with arrows, using the length of the arrows to
indicate the relative magnitude of the forces.
(c) On the diagram at right, draw extended electric field
lines for Q. What feature of these lines indicates the
relative magnitude of the fields at A and B?
D
Electric fields—and hence magnitudes of forces on test
charges—are strongest where field lines are most dense.
So, it is clear from the field lines that for test charges
placed at the different points, |FB| > |FA| > |FC| > |FD|.
B
Q
C
A
(d) Calculate the magnitude of the electric field at point A, in terms of given variables.
E A kQ / rA2 rˆ
If Q is doubled, what is the new value of the electric field at A? If q is doubled instead, what is the new
value of the field at A?
If Q is doubled, E 2 E
If q is doubled E E
1
Discussion Question 2A
P212, Week 2
Electric Field Lines
(e) The region in the box at right contains a uniform
electric field, E. A test charge q is placed in the box.
Compare the magnitude of the field at points A and B.
Write an expression for the force the test charge feels at
point A.
EA
EB
FA
qE
(f) The region in the box at right contains a non-uniform
electric field. Compare the magnitudes of the field at A,B,
and C. If test charges of equal magnitudes (and masses) are
placed at B and C, which one is pushed out of the box
first?
E A EB EC
The charge at B is pushed out of the box first.
2
E
A
E
C
B
B
A
Discussion Question 2B
P212, Week 2
Electric Field Due to Point Charges
Four point charges 2q, q, q, and q are placed at the corners of a rectangle of dimensions a and 3a
as shown in the figure. A fifth charge Q is placed at the center of the rectangle. Our task is to
compute the electric field at the center of the rectangle, and then determine the force on Q.
3a
q
2q
y
a
x
Q
q
q
(a) Each of the 4 charges at the corners contributes to the electric field E at the center of the
rectangle. You’ll have to add these contributions by components. First, start by finding the
magnitudes of all the electric field contributions you will need to add together.
| Eq | = kq/r2
r=
| E2q | = 2kq/r2
5
a
2
Did you draw a sketch? Without it this problem is
much harder than it needs to be …
(b) Without using your calculator, calculate sin( , cos( and tan(
defined in the diagram. (Express your answer algebraically.)
sin = 1/ 10
cos = 3/ 10
where
is the angle
tan = 1/3
(c) Now compute the x- and y-components of the relevant contributions to the electric field at
the center of the rectangle.
Ex =
6kq
Ey =
2kq
5 10 a 2
5 10 a 2
(d) What is the total electric field E at the center of the rectangle, given the particular values q =
3 C and a = 2 cm?
| E | = 2.7 * 107 N/C
(e) Finally, what is the force on the fifth charge Q due to this electric field, given Q = 4 C?
Remember that force is also a vector, and you should give both its x and y components.
Fx = -102 N
Fy = -34 N
Discussion Question 2C
P212, Week 2
Forces Between Point Charges
Consider two charges Q placed at fixed positions a horizontal distance d apart, as shown in the figure.
A third particle of mass m and unknown charge q is then placed a vertical distance h above the fixed
charges. The position of this third particle is arranged so that it is suspended in mid-air above the
other two! This equilibrium condition occurs because the electrical and gravitational forces on the
charge precisely cancel each other.
(a) Calculate the charge q that is required to balance the third particle in mid-air.
Express your answer in terms of the given parameters plus physical constants.
q, m
2
kQq
(d / 2)2 h 2
q
h
(d / 2)2
mg
(d / 2)2
2KQh
h2
h2
mg
0
h
3/2
Q
d
(b) Does your answer make sense? Find three limiting cases where you know what the answer
should be and make sure your expression for q gives the right result. And here’s another important
check: does the sign of q make sense?
Want q and Q to have same sign to insure repulsion.
As h 0 : q
As d
0: q=
k (2Q )q
h2
As m 0 : q 0
1
mg
q
mgh 2
2KQ
Q
(c) Equilibrium situations like the one you have analyzed can be described as either stable or unstable.
Stable equilibrium means that the system will return to its equilibrium configuration when disturbed
by a small amount (like a swinging pendulum). Unstable equilibrium describes situations like a ball
balanced on top of a hill – the slightest disturbance will drastically affect the system. Think about your
suspended charge problem for a moment ... what sort of equilibrium condition is present? Take your
best guess before you turn to the next page.
No restoring force in
z (in-out of page) thus unstable equilibrium
Stable equilibrium means that if you “bump” the charge, the system will provide a restoring force to
bring it back to its resting place. In the opposite case of unstable equilibrium, the system will
encourage the displacement, forcing the charge further away from its equilibrium spot. So let’s try
bumping our suspended charge in different directions …
Note: for the graphs below, we are taking the equilibrium height h of the charge to be 1 meter. 1
(d) Is the balanced charge stable against displacements in y?
y
q, m
h
Q
Q
x
If you bump the charge in the y direction (up or down), will the
combined forces of gravity and electricity provide a restoring force, or will they push it further away?
Check your physical intuition, then consult the plot at right which shows the vertical component Fy of
the force as a function of y. (Note how the force goes to zero at the equilibrium spot y = h = 1 m, as it
must … but here we’re concerned with the behavior of Fy around this point.)
1
For the record, the other parameters used to produce these plots are d = 2 m, m = 3 g, Q = +2.5 C, and q = 1.85 C.
2
Fy
const y for an effective spring. This is what it looks like
locally about the equilibrium point
stable
(e) Is the balanced charge stable against displacements in x?
Again, check your intuition, then the plot. This time it shows the
horizontal component Fx of the force as a function of x (while y is kept at
h).
y
q, m
Fx
const x for an effective spring
h
stable
Q
Q
x
So are we done? ………… not quite, remember that our charges live in 3-dimensional space!
(f) Is the balanced charge stable against displacements in z?
Now the plot shows Fz versus z … does this plot show a restoring force?
Does it make sense?
y
Fz
const z
q, m
unstable
h
Q
Q
x
z
3
Wrapup:
If all went well on the previous page, you found that your suspended charge is stable against
displacements in the x and y directions, but not in z. It’s challenging to think in 3 dimensions! We
worked with force plots on the previous page, but the potential energy map U(x,y,z) of a system is a
much better way to visualize its behavior. Remember from Physics 211 how a system always wants to
minimize its potential energy? That means that wells in the potential energy distribution indicate
places of stable equilibrium – places where the system wants to be. Below are three plots showing the
map U(x,y,z) for your suspended charge (you’ll find out how to calculate them next week). Notice
how clear the conditions of stable and unstable equilibrium are!
4
Discussion Question 2D
P212, Week 2
An Electrostatic Quadrupole
The figure below shows a head-on view of four
long, parallel charged rails (they extend both into
and out of the page). The side rails carry a positive
charge per unit length + . The top and bottom rails
have the same charge density, but with opposite
sign (i.e. they are negatively charged). This
arrangement forms an electrostatic quadrupole. In
the questions below, you will discover why it is
useful in devices which contain beams of moving
charged particles. We will not perform any
calculations in this question … all you need is your
physical intuition.
y
-
rods extend into
and out of page
d
+
+
d
x
-
(a) Sketch the electric field lines in the interior
region of the quadrupole.
Lines go from positive to negative poles.
(b) Given your sketch, where is the electric field the strongest? Where is it the weakest?
Strongest at any pole, weakest at center.
(c) Now imagine that a beam of moving electrons enters
the quadrupole, with velocity parallel to the charged rods.
When the beam enters the device, its cross-sectional profile
is circular. What will the beam look like when it exits the
quadrupole on the other side? (Assume that the force
exerted on the electrons is not too large.)
y
+
beam of electrons
heading into page
+
x
The beam of electrons would be flattened into a oval that is
long in the x-axis and short in the y-axis.
(d) The cathode ray tube (CRT) in a TV monitor contains a beam of electrons directed towards a
fluorescent screen. The screen lights up where the beam hits it. If you were an electrical engineer
designing such a CRT, why might you find it useful to pass the electron beam through a quadrupole
before it hits the screen? (A little hint: quadrupoles are usually used in pairs … can you figure out
why?)
The quadrupole acts to correct errors in the electron firing gun. Non-perfect electrons will still hit pixel.