FREQUENTLY ASKED QUESTIONS
September 4, 2012
Administrative Questions
Will solutions to the homework problems be posted?
Yes, one of the recitation instructors will post solutions each week– look on
Sakai. Please check all your answers using these, and if there’s anything you
don’t understand in the solutions, please come and ask!
Content Questions
How much math are we supposed to know?
You should know basic algebra, geometry, trigonometry. You should know
basic derivatives and integrals. On a test you’ll be given tricky integrals,
but you should know them for things like trig functions and polynomials,
and know to do changes of variable. Some of you might not have seen some
things for a while– if this is the case, you might want to look at Appendix B
of the text to refresh your knowledge.
What’s the difference between dimensions and units?
The “dimensions” of a quantity refer to the basic nature of the quantity, i.e.
how the quantity is related to length, mass, time, charge. Dimensions are
a more abstract, general kind of thing than units; “units” refer to specific
ways of reporting a quantity. A particular quantity can be reported in many
different kinds of units, but it will always have the same dimensions. For
example, force (which can be expressed F = ma) has dimensions of mass ×
length/time2 . Force can be expressed in different units: Newtons, ergs, or
pounds-cm per square hour.
Checking the dimensions your answers is a very, very useful habit to get
into: if they don’t come out to what you expect, you know you’ve made a
mistake somewhere.
Can you explain dimensionality?
(Note the difference between “dimensions of a quantity”– see above question–
and the “dimensionality of a system”).
In the context of our homework problems, the “dimensionality” of a system refers to how many variables you need to integrate over. If you only need
to integrate over one variable, it’s a 1D problem; if you need to integrate over
a surface, it’s a 2D problem; if you need to integrate over a volume, it’s a 3D
problem. All real-world systems are really 3D, but sometimes they can be
treated as 1D or 2D if one or more dimensions are unimportant, or if they’ve
already been integrated over. (See also the question below about finding dq.)
How is finding the electric field for a continuous distribution the
same as finding the force?
Electric field and Coulomb force are not the same quantity, but they are
related. The electric field from a charge distribution is defined as the force
~ is much like
on a positive test charge q divided by q. So determining E
~
finding F on q, but with a factor 1/q. So the techniques for finding electric
field for a continuous charge distribution can be used in exactly the same way
as for finding the force on a specific charge q0 from that charge distribution–
the only difference is that you’ll have an extra factor of q0 in your force
expressions.
What is dq? Change in amount of charge?
In these problems, the meaning of dq is an infinitesimally tiny chunk of
charge. When treating continuous charge distributions, we’re cutting them
up into imaginary pieces with charge ∆q. As the pieces get infinitesimally
small, and the number of them goes to infinity, ∆q → dq.
How do you know how to find dq so as to do the integral?
The key thing for doing these force and field integration problems is often
figuring out how to turn a charge element dq into a space coordinate you can
integrate over: this is step 8 of the how-to.
Usually the dimensionality of the system (whether it’s 1D, 2D or 3D)
will help you. Sometimes the dimensionality is obvious (for example a line
charge is 1D), but sometimes you can use what you already know to reduce
the dimensions and make the problem easier. To find the dimensionality,
think about what you are integrating over: if just one variable, it’s 1D; if
two, it’s 2D and so on.
Applying to this week’s examples:
• For the circular arc of charge: this is a 1D problem, and linear charge
density is given as λ(θ), a charge density that varies along the arc.
Here, a length element is ds, which is a tiny arc length. In this case
you use dq = λds, and you need to remember if you integrate over a
polar coordinate θ, you have ds = Rdθ.
• For the hollow cylinder: this looks at first glance like a 2D problem,
since it’s a surface you want to add up field contributions over. However, as suggested in the hint, you can use a previous result to turn
it into a 1D problem, where you just integrate over x. From example
23-7 in the text, you know the field from a ring as a function from
distance to the center of the ring, the radius of the ring, and the charge
kx
on the ring E = (a2 +x
2 )3/2 Q. Now, you can consider the tube to be
sliced into many rings of infinitesimal width dx, each with charge dq.
So the contribution of one of these skinny rings a distance x away is
kx
dE = (R2 +x
2 )3/2 dq. Now, you want dq in terms of dx. You know the
total charge and total length of the cylinder, and dq/Q = dx/l, and
dq = Qdx/l (or dq = λdx, where λ = Q/l, the linear charge density).
Now you can plug that in and crank the integral over x. So using what
you know about the field from a ring, you’ve turned the problem into
a 1D integral.
• For the solid cylinder (which we didn’t get to, but which you should do
and check your answer from posted solutions): you know the field from
a disk, from example 23.8. Now you can treat the cylinder as a stack of
thin disks, each of charge dq, and again relate dq to dx by dq = Qdx/l.
It takes a bit of practice to determine the best approach, and sometimes
you might write down a couple of things before finding the way that works.
What exactly is λ?
Linear charge density λ is charge per unit length for a continuous charge
distribution. If the charge distribution is uniform (not varying along the
length), you can find λ from λ = Q/L, where Q is the total charge and L
is the total length. If the distribution is not uniform (like in the half-circle
problem of this week), you can’t use that and you need to integrate over the
length coordinate.
How did you find λ0 in the half-circle loop of charge problem?
R
We related λ0 (not given) to Q (given) by using that Q = dq. In this problem λ = λ0 cos θ is given– it’s a non-uniform charge distribution. The coordinate to integrate is θ, and dq = λds, where ds is an infinitesimal
arc length
R π/2
of the wire; since ds = Rdθ, we have the integral Q = −π/2 λ0 R cos θdθ. This
gives the relation Q = 2λ0 R, and hence λ0 = Q/(2R), which can be used in
the overall answer.
Why did you need a cos θ in the half-circle loop of charge problem?
By symmetry, we know we can ignore the x-components of all the force
contributions, because the charge distribution is left-right symmetric. So
we need to add up the y-components of all the forces due to dq elements
along the wire. These forces all point to the left or right and not straight
down. The magnitude of each force contribution is dF = kq0 dq/R2 , but the
y-component is only the part of the vector going straight down, and that has
a factor cos θ (it varies as you go around the half-ring).
The force integral actually has two cos θ’s: one for the y-component, and
one from the line charge density.
Where did you get the π in the half-circle loop of charge problem?
This came from the integral
R π/2
−π/2
cos2 θdθ, which comes out to π/2.
Why is λ = Q/l in the example question?
I guess you are referring to the cylinder: see the answer to the question above
about finding dq. In this particular example, you can use a previous result
that tells you the electric field from a ring as a function of distance and
charge of the ring. You can just add up fields from infinitesimally ring slices
of charge dq to get the total field from the cylinder. There is now only one
dimension to integrate over, x. To relate dq to the x coordinate, dq = λdx,
where λ is charge per length. Since the cylinder is uniform, you can write
λ = Q/l: charge per length is total charge over total length.
You can also think of it as follows: the ratio of length dx to total length
.
l is the same as the ratio of charge dq to total charge Q, i.e. dxl = dq
Q
Will we ever use ρdV for continuous charge distributions?
I am not sure exactly what problems will be assigned, but probably yes.
Why do we use field lines?
They are a useful visualization tool for electric field– the denser the lines,
the higher the field.
Joke of the Week
A bright young boy looks over to his left to realize he’s been sitting next to
none other than Albert Einstein. Excitedly, he asks Einstein, “Sir, do you
happen to know if Durham stops at this train?”