I'm studying for a course in electromagnetism, and I've been given an electric field for which I need to find the associated scalar potential. The field is the field generated by a sphere of radius $R$ with constant charge density $\rho$ throughout its volume, so that the total charge $Q=\dfrac{4\pi r^3 \rho}{3}$contained in the sphere is constant.

The electric field is given by $\vec{E}_{\text{in}}(\vec{r})=\dfrac{Q}{4\pi \epsilon_0 R^3}r$ and $\vec{E}_{\text{out}}(\vec{r})=\dfrac{Q}{4\pi \epsilon_0 r^2}$, where the former is valid for $r\leq R$ and the latter for $r\geq R$. This I've calculated before and I do not have trouble with. The scalar potential $\phi(\vec{r})$ is defined by $\vec{E}=-\vec{\nabla}\phi$. The provided solutions to the problem are hand written but I'll type them here using the exact same notation:

$\phi_{\text{in}}=-\int \vec{E}_{\text{in}}d\vec{r}=-\dfrac{Qr^2}{8\pi \epsilon_0 R^3} + C_1$

$\phi_{\text{out}}=-\int \vec{E}_{\text{out}}d\vec{r}=\dfrac{Q}{4\pi \epsilon_0 r} +C_2$

This is literally all the information I've been given. I really don't know what these integrals are, nor how they follow from the above equation. I can see that the result of the first integral for example is just the indefinite integral $-\int \dfrac{Q}{4\pi \epsilon_0 R^3}r dr$ but I can't see how this stage was reached. I think my professor intended this to be a $\cdot$ in the integrals but has missed them out. Even so I can't figure out where these integrals come from (i.e. why they give the potential), what these integrals mean, or (if they are indeed surface integrals) how to evaluate them.

Any clarification would be much appreciated!

  • $\begingroup$ Possible duplicate of this question. In that third equation you probably mean $R^2$ rather than $R^3$ ... don't you? The integrals are the inverted form of $E=-\nabla\phi$. They are not surface integrals, they are line integrals. You must have missed a chunk of the theoretical development for some reason. Do you have a textbook or notes that might fill in the gap? $\endgroup$ – garyp Jun 27 '14 at 2:09
  • $\begingroup$ Huh, I asked the same question in the maths stack exchange and was told it was a surface integral! Well yeah I definitely am lacking in the knowledge of the theory behind this, but it's not in my course notes despite this being a question from said course, unfortunately. $\endgroup$ – James Machin Jun 27 '14 at 2:19
  • $\begingroup$ @garyp Also, how would I calculate these line integralas without knowing what the curve is? $\endgroup$ – James Machin Jun 27 '14 at 3:10
  • $\begingroup$ @BMS has the answer to that. But if this is for an E&M course, I'm concerned that this is a question at all. This is basic, important stuff that should be well covered. If not in the class, then certainly in a (any) textbook. Something is amiss with the lectures, or notes, or textbook. Take this as a warning that you may be in for extra work going forward! $\endgroup$ – garyp Jun 27 '14 at 11:53
  • $\begingroup$ @garyp Yeah my calculus is really not up to scratch but I don't mind learning the fundamental prerequisites to the course in conjunction with the actual course content - it makes the calculus less abstract. I should add that I usually can calculate line integrals, it's just the method I learned to calculate line integrals uses the parametrization of the curve, (i.e. I'd do $\int \vec{E}\cdot \vec{r} '(t) dt$ where $\vec{r} (t)$ is the parametrization of the curve. $\endgroup$ – James Machin Jun 27 '14 at 12:51

In addition to BMS answer, I want to point out the integration part as I have seen,in the comments, you have some problems in the integration part.

First you should have written the unit vectors in the expression of the electric field.

The electric fields are

$\vec{E}_{\text{in}}(\vec{r})=\dfrac{Q}{4\pi \epsilon_0 R^3}r\hat{r}$ and $\vec{E}_{\text{out}}(\vec{r})=\dfrac{Q}{4\pi \epsilon_0 r^2}\hat{r}$ $\ $(obviously, Electric field is radially outward due to spherical symmetry symmetry)


$\vec{E} \cdot d\vec{r}=-\vec{\nabla}\phi \cdot d\vec{r}= -d\phi$

$\int_\vec{a}^\vec{r}\vec{E} \cdot d\vec{r}=-\int_\vec{a}^\vec{r}d\phi$

$\phi(\vec{r})-\phi(\vec{a})=-\int_\vec{a}^\vec{r}\vec{E} \cdot d\vec{r}$

out of many possible path, we are going to take our path radially , due to conservative nature of electric field. so $d\vec{r}= d(r \hat{r})=dr \hat{r}$, since $\hat{r}$ remains same along the radial direction, we have $d\hat{r}$=0.This would not be case if we would have taken any other path between the points $\vec{a}$ and $\vec{r}$.

so, $\int_\vec{a}^\vec{r}\vec{E_{in}} \cdot d\vec{r}=\int_\vec{a}^\vec{r}\dfrac{Q}{4\pi \epsilon_0 R^3}r\hat{r} \cdot dr \hat{r}=\int_\vec{a}^\vec{r}\dfrac{Q}{4\pi \epsilon_0 R^3}r dr=\int_a^r\dfrac{Q}{4\pi \epsilon_0 R^3}r dr $

see at the last step vector sign has been dropped because the integration depends only on r, but not on ($\theta,\phi$).

so, $\int_\vec{a}^\vec{r}\vec{E_{in}} \cdot d\vec{r}=\dfrac{Qr^2}{8\pi \epsilon_0 R^3} -\dfrac{Qa^2}{8\pi \epsilon_0 R^3}$

so, $\phi_{in}(\vec{r})-\phi_{in}(\vec{a})=-(\dfrac{Qr^2}{8\pi \epsilon_0 R^3} -\dfrac{Qa^2}{8\pi \epsilon_0 R^3})=-\dfrac{Qr^2}{8\pi \epsilon_0 R^3} +\dfrac{Qa^2}{8\pi \epsilon_0 R^3}$ $\phi_{in}(\vec{r})=-\dfrac{Qr^2}{8\pi \epsilon_0 R^3} +\dfrac{Qa^2}{8\pi \epsilon_0 R^3}+\phi_{in}(\vec{a})$

$\phi_{in}(\vec{r})=-\dfrac{Qr^2}{8\pi \epsilon_0 R^3} +C1$

where C1=$\dfrac{Qa^2}{8\pi \epsilon_0 R^3}+\phi(\vec{a})$

Note: I think you have some problem with vector analysis. You can look at the "Vector Analysis (Schaum's Outline) by Spiegel". It is an excellent book.

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I really don't know what these integrals are

This issue seems very broad, and I don't know what is really meant by this statement. Maybe the info below will help.

I think my professor intended this to be a ⋅ in the integrals but has missed them out.

Agreed. Just look at the definition of the electric potential to see this.

I don't know [...] how they follow from the above equation.

I'll briefly outline the general steps. The nitty gritty is for best left for a more focussed question.

We want to evaluate the line integral $$\phi\equiv-\int \vec E \cdot d \vec s$$ along some path, which is typically labeled $C$. (You'll see subscripts on the integral indicating this sometimes, like $\int_C$.) A common question when learning this is to ask which path? The answer is any path from an arbitrary reference point (that you choose) to the point $(x,y,z)$ at which you want to evaluate the potential. Often people choose infinity to be the reference point, meaning physicists would typically write something like $r=\infty$ as the lower bound. Now, you don't have to make that choice. If you make a different choice for the arbitrary reference point, and then compare your final result for the potential with the result from yet another choice, you'll find that the results differ only by a constant value.

That's pretty amazing. (Try it to convince yourself.) Now, a really clever way of doing these line integrals for potential is to just write out the indefinite integral and slap on the integration constant on the end. That's what your instructor did.

Another way of seeing why this works is to remember that, in the end, we're typically more interested in the electric field $\vec E=-\vec \nabla \phi$. So who cares what the integration constant is? It goes away when you take the derivative.

I can't figure out where these integrals come from (i.e. why they give the potential), what these integrals mean

They come from the definition of the electric potential $\phi$. Nothing more. The phrase electric potential means $-\int \vec E \cdot d\vec s.$

As for what such an integral means, not much. But differences in electric potential (aka voltage or electric potential difference) tell you something akin to the potential energy per unit charge. There are plenty of resources online or in your textbook elaborating on this point.

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