# E-field questions

#### leright

ok, so I have a physics test monday, and I know the prof will make us calculate the e-fields, and voltages everywhere is space produced by various objects with uniform charge densities using Gauss's law. So, he wants us to derive a general expression that describes the e-field and voltage everywhere in space around a charged object, say, as a function of the radius....he also will want us to determine the capacitance due to a certain voltage between two oppositely charged objects. ok, so that isn't too bad, right?

Well, I know he is going to give us a problem where we have a solid sphere of uniform charge density (+) and this solid sphere is surrounded by a spherical shell with uniform surface charge density (-). He will want us to determine the e-field everywhere (meaning, inside the inner solid sphere, in between the inner solid sphere and outer shell, and outside the outer shell.

I have trouble seeing how the fields from the outer, middle, and inner regions cancel one another out or add to one another. Inside of the solid sphere, the fields cancel out. In between the outer shell and the inner solid sphere it seems like the fields would add, since they are pointing in the same direction, right? Finally, outside the outer shell, it would just be the field due to the shell, right? no charge would contribute to the outer shell's field, other than the outer shell charge, correct? I just want to ensure that my line of thinking is correct here.

Also, we have only talked about electric fields so far, but I am aware that magnetic fields are a different phenomenon...can you kinda explain the difference to me? From what I understand, electric fields are simply forces produced by stationary charges (where the efield is measured in force/coulomb), and these fields can facilitate work on other charged objects. As I understand it, magnetic fields are a different ype of field, but related to electric fields, that is due to charges in motion, right? Can someone elaborate on this for me?

Also, how do e-fields and m-fields tie in with electromagentic fields. I am just horribly confused by all of this stuff lately.....it is making me depressed. :(

Also, what exactly are electromagnetic waves? Are they just electromagentic fields?

bah...

Oh, and the prof said he is putting a quadrupole problem on the test, which consists of determining the field produced by an arrangement of equal charges in the following order: + - + -. The charges are on the same axis. The distance between the charges is 'd'. He wants us to find the e-field at a point a distance 'a' from the first charge, where a >> d. the e-field point lies on the same axis as the 4 charges. I would just reason that since a >> d then the fields due to the charges are about the same magnitude and cancel out, since a ~= a + d ~= a + 2d ~= a + 3d. However, they don't QUITE cancel out and a more accurate answer can be determined....HE (evil bastard) wants us to use the binomial theorem to determine the e-field. :grumpy: Can somone explain how this problem would be solved? This leaves me very confused.....he did it in class, but I missed part of the lecture (yes, I go to 99% of my lectures...). I went to him and asked for his help, and he was quite useless, and nobody else in the class really knows how to do it either...There's a 99% chance this problem will be on the test. I talked to a physics tutor the other day and he was having trouble with the problem too.

Any help with any of these problems would be greatly appreciated.

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#### OlderDan

Homework Helper
Every spherically symmetric charge distribution must have an electric field that is radial, pointing toward or away from the center of symmetry. Therefore, the normal component of the electric field for any concentric spherical surface is constant. It follows that there can be no contribution to the electric field at any radius from the charge distribution that is at a greater radius from the center of symmetry. For your solid sphere surrounded by concentric shells of charge you need not worry about anything outside the radius of interest. The field at any given radius has constant magnitude and depends only on the total charge enclosed by the spherical surface of that radius. You must consider all of the enclosed charge, not just the outermost shells. When the spherical symmetry is broken, then (and only then) charges at greater radius can contribute.

In classical electrodynamics electric fields arise from charges, and magnetic fields arise only when charges are moving. Whether a charge is moving or stationary depends on the frame of reference from which it is viewed. If you move along with a charge, you see it at rest and experience only electric fields. If it is moving relative to you, a charge will have an associated magnetic field.

Electromagnetic fields can exist in empty space as electromagnetic waves because time varying magnetic fields induce electric fields and time varying electric fields induce magnetic fields. The fields originate from charges, but you don't need the charges to be anywhere nearby for the time varying electromagnetic field to propagate through space as a wave.

#### OlderDan

Homework Helper
For the quadrapole problem, the distance from each charge to the point of interest can be written as you have done. Since the electric field from each charge is inversely porportional to the distance squared, you will have to add inverse squares that are slightly different. For the terms that involve d you will have

$$\frac{1}{(a+nd)^2} = \frac{(1+\frac{nd}{a})^{-2}}{a^2} \approx \frac{(1 - \frac{2nd}{a})}{a^2}$$

When you add all the terms together the contributions from the "1"s in the numerator will cancel and you will be left with a sum involving terms that are each inversely proportional to $$a^3$$

#### leright

OlderDan said:
For the quadrapole problem, the distance from each charge to the point of interest can be written as you have done. Since the electric field from each charge is inversely porportional to the distance squared, you will have to add inverse squares that are slightly different. For the terms that involve d you will have

$$\frac{1}{(a+nd)^2} = \frac{(1+\frac{nd}{a})^{-2}}{a^2} \approx \frac{(1 - \frac{2nd}{a})}{a^2}$$

When you add all the terms together the contributions from the "1"s in the numerator will cancel and you will be left with a sum involving terms that are each inversely proportional to $$a^3$$
How do you make the final approximation, exactly? This is what I am having trouble with.

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#### OlderDan

Homework Helper
leright said:
How do you make the final approximation, exactly? This is what I am having trouble with.
The binomial theorm

http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html#bt

$$(a+b)^n = a^n + na^{n-1}b + \frac{n(n-1)}{2}a^{n-2}b^2+...+b^n$$

for any n. When a >> b the terms get much smaller very quickly and it is a good approximation to keep only the leading terms. In the quadrupole problem, if you keep only the first term it gives zero, so you keep the first two terms. If you wanted more precision you would keep more, but you are only interested in distances far from the charges, so the first two terms is good enough.

#### leright

OlderDan said:
The binomial theorm

http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html#bt

$$(a+b)^n = a^n + na^{n-1}b + \frac{n(n-1)}{2}a^{n-2}b^2+...+b^n$$

for any n. When a >> b the terms get much smaller very quickly and it is a good approximation to keep only the leading terms. In the quadrupole problem, if you keep only the first term it gives zero, so you keep the first two terms. If you wanted more precision you would keep more, but you are only interested in distances far from the charges, so the first two terms is good enough.

ok, I am just not seeing how the ones end up cancelling though....what is going wrong? I added the e-field expressions due to each of the charges, factored out kQ, and then used the binomial theorem to add the terms up....but the "ones" do not cancel.

Man, I hate this....

#### OlderDan

Homework Helper
leright said:

ok, I am just not seeing how the ones end up cancelling though....what is going wrong? I added the e-field expressions due to each of the charges, factored out kQ, and then used the binomial theorem to add the terms up....but the "ones" do not cancel.

Man, I hate this....
Sure they do. The charges alternate + - + -. You have two plus ones and two minus ones. Be careful with those signs when you add the remaining terms.

#### leright

OlderDan said:
Sure they do. The charges alternate + - + -. You have two plus ones and two minus ones. Be careful with those signs when you add the remaining terms.
THANKS! You're the man.

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