Finding a scalar field whose gradient is a given conservative vector field

Question

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. I was going to originally post this in the physics section but I think my problems are more calculus related. 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.

Any clarification would be much appreciated!

Answer 1

What the OP has given as the electric fields inside and outside the sphere are only the magnitudes of these fields, as has been emphasized in other comments and answers. I suspect that those expressions were derived using Gauss' Law, which relied on spherical symmetry to assume two things: (i) that the field at a point with position vector $\vec{r}$, a distance $r = |\vec{r}|$ from the center of the sphere, depended only on this distance, and (ii) that the field was directed radially outward from the center of the sphere along the unit vector $\hat{r} = \vec{r}/r$. So the electric field vector is a piecewise continuous function of the radial coordinate $r$ alone, with direction $\hat{r}$, given by $$\begin{align}\vec{E}(r) \,=\, \begin{cases}~~\vec{E}_{in}(r) = \dfrac{q}{R^3}\,r\,\hat{r}, ~~~ r \leq R, \\ \\~~ \vec{E}_{out}(r) = \dfrac{q}{r^2}\,\hat{r}, ~~~ r \geq R, \end{cases}\end{align}$$ where I am writing $$q = \dfrac{Q}{4\pi\epsilon_0}$$ for convenience. Since $\vec{E}(r)= -\nabla \phi(r)$ depends only on the radial coordinate $r$, it reduces to $$\vec{E}(r) = -\dfrac{d\phi}{dr}(r)\,\hat{r}.$$ The negative derivative of the scalar function $\phi(r)$ with respect to $r$ thus gives the radial (and only) component of $\vec{E}$, so from the piecewise expressions for the electric field we find the piecewise expressions for the derivative of $\phi$: $$\begin{align}-\dfrac{d\phi}{dr} = \begin{cases}~~\dfrac{q}{R^3}\,r, ~~~ r \leq R, \\ \\~~ \dfrac{q}{r^2}, ~~~ r \geq R, \end{cases}\end{align}$$ The indefinite integral of each side in the two cases gives $$\phi(r) = -\int\dfrac{q}{R^3}\,r\,dr = - \dfrac{q}{R^3}\dfrac{r^2}{2} + C_1, ~~~~r \leq R,$$ and $$\phi(r) = -\int\dfrac{q}{r^2}\,dr = \dfrac{q}{r} + C_2, ~~~~r \geq R.$$ In the second expression with $r \geq R$, in order for the potential to vanish as $r \to \infty$, we must set the integration constant $C_2 = 0$, hence $$\phi(r) = \dfrac{q}{r}, ~~~~r \geq R.$$ In order for the potential to be continuous at $r = R$, the potentials must be equal at the boundary: $$-\dfrac{q}{R^3}\dfrac{R^2}{2} + C_1 = \dfrac{q}{R},$$ from which we find $$C_1 = \dfrac{3q}{2R}.$$ The scalar potential is thus given by the piecewise continuous function $$\begin{align}\phi(r) \,=\, \begin{cases}~ -\dfrac{q}{R^3}\dfrac{r^2}{2} + \dfrac{3q}{2R}, ~~~r \leq R,\\ \\~~ \dfrac{q}{r}, ~~~r \geq R.\end{cases}\end{align}$$ In terms of $q = Q/4\pi\epsilon_0$, these are $$\begin{align}\phi(r) \,=\, \begin{cases}~~-\dfrac{Q}{8\pi\epsilon_0 R^3}\,r^2 + \dfrac{3Q}{8\pi\epsilon_0 R}, ~~~r \leq R,\\ \\ ~~\dfrac{Q}{4\pi\epsilon_0 r}, ~~~r \geq R.\end{cases}\end{align}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \Phi\pars{\vec{r}} & = -\int_{\vec{a}}^{\vec{r}}\vec{\mrm{E}}\pars{\vec{s}}\cdot\dd\vec{s} + \Phi\pars{\vec{a}} = \overbrace{\int_{\vec{a}}^{\vec{r}}{Q \over 4\pi\epsilon_{0}s^{2}}\,{\vec{s} \over s}\cdot\dd\vec{s}}^{\ds{% \left\{\begin{array}{l} \mbox{Note that}\ \\ \ds{\vec{s}\cdot\dd\vec{s} = {1 \over 2}\,\dd\pars{\vec{s}\cdot\vec{s}}} \\ = \ds{{1 \over 2}\,\dd\pars{s^{2}} = \color{red}{s\,\dd s}} \end{array}\right.}}\ +\ \Phi\pars{\vec{a}} \\[5mm] & = -\,{Q \over 4\pi\epsilon_{0}}\int_{a}^{r}{\dd s \over s^{2}} + \Phi\pars{\vec{a}} = {Q \over 4\pi\epsilon_{0}}\pars{{1 \over r} - {1 \over a}} + \Phi\pars{\vec{a}} \end{align}

Set $\ds{\Phi\pars{\vec{a}} = 0}$ as $\ds{a \to \infty}$ such that $\bbx{\ds{\Phi\pars{\vec{r}} = {Q \over 4\pi\epsilon_{0}r}}}$.

Answer 3

The integrals are over the surface $S$ of the sphere (at radius $R$). The confusion might be because you overlooked a $\cdot$. The integrand is the dot product of $\vec{E}$ with the unit normal to the surface, which your professor has written as $d\vec{r}$ but wouild more commonly be written as $dS$. The integral can be read in spherical coordinates as $$ \int_S E_r(R, \theta, \phi) r\,d\theta\,d\phi $$ and the $\vec{E}\cdot \vec{r}$ is just expressing that you need to take the $r$ component of the field.