Electric Field inside uniformly charged dielectric sphere

[Old Guy]

Homework Statement

Either Coulomb's Law or Gauss' Law and r>R, I get $$\textbf{E}=k\frac{Q}{r^{2}}\textbf{r}$$. Q is the total charge. If R is the radius of the sphere, shouldn't the field at a point a<R simply be $$k\frac{Qa}{R^{3}}$$? Inother words, the field is proportional to the enclosed charge, and the enclosed charge is proportional to the ratio of the radii cubed? Griffiths' answer is $$k\frac{Q}{R^{3}}\textbf{r}$$, i.e., there is no dependence on a. And it appears to be dimensionally inconsistent without a.

Homework Equations

The Attempt at a Solution

 

[kuruman]

What is the statement of the problem?

 

[Old Guy]

Given a sphere of radius R with a uniform volumetric charge density $$\rho$$, find an expression for the electric field at an arbitrary point inside the sphere.

 

[kuruman]

What you call a, Griffiths calls r.

 

[Old Guy]

Yes, of course. I'm not working the Griffiths problem, but the nature of the problem is exactly the same. Whether you call it a or r, I believe it should appear in the numerator in any case. If it doesn't, it would say that the field is independent of the enclosed charge, which violates Gauss' Law (and, as I mentioned, is dimensionally inconsistent).

 

[kuruman]

Is there a question here? The expression

$$E =k\frac{Q r}{R^{3}}$$

as derived from Gauss's Law is the correct answer to the statement of the problem that you have provided. It is a dimensionally correct expression.

 

[Old Guy]

Thank you, that is what I was looking for. Griffiths' solution manual, and several other sources which might have been taken from there, give the answer as E=k$$\frac{Q}{R^{3}}\textbf{r}$$ (note there is no r in the numerator) and I just wanted some assurance that I hadn't made some fundamental mistake in what should be a pretty basic problem. Thanks again.

 

[gabbagabbahey]

Thank you, that is what I was looking for. Griffiths' solution manual, and several other sources which might have been taken from there, give the answer as E=k$$\frac{Q}{R^{3}}\textbf{r}$$ (note there is no r in the numerator) and I just wanted some assurance that I hadn't made some fundamental mistake in what should be a pretty basic problem. Thanks again.

You don't seem to understand the notation Griffiths is using...$\textbf{r}\equiv r \mathbf{\hat{r}}$ is a vector, with magnitude $r$ and pointing radially outwards.

 

[Old Guy]

Doh! How embarrassing! You are right, of course, I can see it now. Thank you.