I have to solve the following integral $$\Phi(\boldsymbol r) =- \int_{\boldsymbol r_0}^{\boldsymbol r}\boldsymbol{E}(\tilde{\boldsymbol r}) \cdot d\boldsymbol s$$ This comes from Electrostatics (I'm computing the electric potential of a charge Q at the origin, with reference point $\boldsymbol r$ being asymptotic infinity) We know that in general for a charge at the origin$$\boldsymbol E(\boldsymbol r) = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{\boldsymbol e}_r$$

My trial

Clearly we use polar coordinates $\{r, \theta, \phi\}$ because of the spherical symmetry.Then the line element becomes $d\boldsymbol s = dr\hat{\boldsymbol e}_r+rd\theta\hat{\boldsymbol e}_{\theta}+r\sin{\theta}d\phi\hat{\boldsymbol e}_{\phi}$ so then $$-\int_{\boldsymbol r_0}^{\boldsymbol r}\boldsymbol{E}(\tilde{\boldsymbol r}) \cdot d\boldsymbol s = -\frac{Q}{4\pi\epsilon_0}\int_{\boldsymbol r_0}^{\boldsymbol r}\frac{1}{\tilde{r}^2}\hat{\boldsymbol e}_{\tilde r}\cdot d\boldsymbol s =-\frac{Q}{4\pi\epsilon_0}\int_{\boldsymbol r_0}^{\boldsymbol r}\frac{1}{\tilde{r}^2}d\tilde{r}$$

Because the spherical polar basis is an orthogonal basis, hence they are all perpendicular to each other and the result of the dot product is $1$ coming from the radial vector. But now I have a scalar integral and I will have to evaluate the result at a vector?

Can someone explain to me (possibly explaining each passage and with some strong mathematical rigor) how someone would finish this calculation?


Note that in general we have

$$\Phi(\vec r)=\Phi(\vec r_0)-\int_{\vec r_0}^{\vec r} \vec E(\vec r')\cdot d\vec \ell'$$

The notation for the line integral can be more explicitly written by parametrizing the curve traversed from $\vec r_0$ to $\vec r$. Let the curve be described parametrically as $\vec r(t)$ where $\vec r=\vec r(1)$ and $\vec r_0=\vec r(0)$. Then, we can write

$$\int_{\vec r_0}^{\vec r} \vec E(\vec r')\cdot d\vec \ell'=\int_0^1 \vec E(r(t),\theta(t),\phi(t))\cdot \left(\hat r\frac{dr(t)}{dt}+\hat r\frac{d\theta(t)}{dt}+\hat r\frac{d\phi(t)}{dt}\right)\,dt \tag 1$$

If $\vec E$ has only a radial component that depends on $r(t)$ only, then $(1)$ becomes

$$\begin{align} \int_{\vec r_0}^{\vec r} \vec E(\vec r')\cdot d\vec \ell'&=\int_0^1 E_r(r(t))\frac{dr(t)}{dt}\,dt\\\\ &=\int_{r(0)}^{r(1)}E_r(r)\,dr \tag 2 \end{align}$$

Note that $r(1)=|\vec r|$ and $r(0)=|\vec r_0|$. Thus, we can write the integral in $(2)$ as

$$\int_{\vec r_0}^{\vec r} \vec E(\vec r')\cdot d\vec \ell'=\int_{|\vec r_0|}^{|\vec r|} E_r(r)\,dr \tag3$$

Finally, using $E_r=\frac{Q}{4\pi \epsilon_0 r^2}$ we find from $(3)$ that

$$\Phi(\vec r)=\Phi(\vec r_0)+\frac{Q}{4\pi \epsilon_0 |\vec r|}-\frac{Q}{4\pi \epsilon_0 |\vec r_0|}$$

Letting $|\vec r_0|\to \infty$ and setting $\Phi(\infty)=0$, we obtain the coveted result

$$\Phi(\vec r)\frac{Q}{4\pi \epsilon_0 |\vec r|}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.