Expert: Daniel Mazur - 4/7/2010

Question
QUESTION: Hi, I'm a Junior taking a regular Physics class because my high school does not have any AP physics classes available. I am though Learning AP physics C on my own but i have run into a small problem and I have no one to ask for help. I tried posting this in the physics-physics section but my question didn't get answered.

My question is about the application of Gauss' law to a sphere with a symmetric charge distribution that depends only on the radius. In trying to find the electric field at a point inside the sphere the Gaussian surface would be a sphere with radius r less than the radius (a) of the actual sphere (so r < a). My book suggests considering thin spherical shells of radius R and thickness dR.


Now have e*integral(E*dA)=q(charge enclosed by Gaussian surface)
and the charge enclosed is q=integral(p*dV)
so we now have E*A=integral(p*dv)/e

(side not not important to my problem but p=(p(0))*(r/a)^2, (a) being the radius of the actual sphere)

where E is the electric field, A is the surface area of the Gaussian surface, p is the function for the volume charge density, dV is the volume of the spherical shells, and im using e to represent the electric constant.


My problem comes from solving for dV, normally for the volume of a spherical shell i would take the volume using the outside surface radius and subtract the volume found using the inside surface radius,4/3(pi)R^2-4/3(pi)*(R-dR)^2. The book though gives this dV=4(pi)R^2*dR.

my question is how is this correct? I understand calculus and physics pretty well on their own but when i put them together i get confused, so this might turn out to be more of a problem with my calculus but i would appreciate it if you could still help me.



Thanks.

ANSWER: Dear Carlos,

your question is a fair one and once upon a time I was asking a similar one. The answer is that for infinitesimal dR (in the dR->0 limit) the correct formula equals the one you are suggesting - that is CORRECTED for a typo you have made, inadvertently, by writing ^2 instead of ^3. If you then expand the CORRECTED formula, you will get the following:
4/3(pi)R^3-4/3(pi)*(R-dR)^3 = 4/3(pi)R^3-4/3(pi)*R^3+4(pi)R^2*dR-4(pi)R*dR^2+4/3(pi)dR^3 = 4(pi)R^2*dR-4(pi)R*dR^2+4/3(pi)dR^3
In the last expression there are 3 terms, one proportional to dR, second to dR^2 and third to dR^3. Study of the differential calculus tells us, that it is enough to keep the term linear in dR, since when we integrate once over dR, the linear term already gives a (generally) non-zero result, whereas the quadratic and cubic terms remain zero. Differential calculus is a largely a linear maths discipline... and so, 4(pi)R^2*dR is correct.

Take care!

Daniel

---------- FOLLOW-UP ----------

QUESTION: Oh I hadn't realized i put ^2 instead of ^3 that's embarrassing. Anyway Thanks for the help I understood all of what you said except for the last part. What does it mean when we have dR cubed or squared does it imply that we must integrate twice or three times to get rid of those? I realize this is not necessary but is that what it means?

also on a side note I really enjoy physics and i want to study it in college, I am pretty sure i want to go into theoretical physics. So can you recommend good colleges in the U.S. preferably in California for this type of major? I know Stanford is a good one but i don't really know of any others.

Answer
Hi Carlos,

the best answer I can give to your question
[Q]What does it mean when we have dR cubed or squared does it imply that we must integrate twice or three times to get rid of those?[/Q]
is that if the term with dR^2 or dR^3 stood alone (!) for our consideration, we would indeed have to subject it to double or triple integration, respectively, in order to get finite (i.e. "in principle" not identically zero as indefinite integrals) terms from them.

It looks strange of course, but look at dR as a difference (usually labeled "deltaR") instead of a differential. If you draw a sketch, you will clearly see, what bits of the volume of the spherical shell are represented by the dR^2 and dR^3 terms. We will observe that only one dR is actually a length element in radial direction. The other two are in directions perpendicular to that, they represent R*sin(Theta)*dTheta and R*dPhi in spherical coordinates.

Suppose now that you've modified the terms for the use of dTheta and dPhi. If you consider the integrations over these angles correctly, they will still give you zeros. Why? Because of the correct LIMITS of the definite integrals. In the dTheta integral Theta must go from 0 to Epsilon, where Epsilon->0. If both limits of integration are the same (like 0 and 0, or 11 and 11, or any other equal values), integral is identically zero. Exactly the same happens to Phi - the limits of integration are essentially the same, so in the limit of differential calculus those integrals stay zero.

I hope this reasoning is transparent enough, may it serve you well.

Daniel