# Electric field at the focus of a parabolic line

Today my physics teacher was teaching me magnetism and application Biot–Savart law ;I came to realisation that Biot–Savart law is some what similar to coloumbs law.

While teaching us application of Biot–Savart law he taught us how to find magnetic field in a thin wire,in a cylinder and etc....

One of the problem he gave us was

"Consider a wire, bent in a shape of a parabola, kept in XY plane with focus at origin. The disntace from apex to focus is d. The wire carries current I. Find the magnetic field at origin"

which we were able to do it..

But then a simple question struck to my mind

"To determine the electric field at the focus of a parabolic line of uniform charge density."

i asked my teacher about it ...he thought for a while but was unable to give me a satisfactory answer or a process to proceed the problem.

can anyone tell me a method to find a solution to my thinking?

• This is a homework-like question and users must avoid to give full solutions in their answers. Aug 7, 2018 at 4:56

Edited to leave answer out

First, let's assume we have a downward facing parabola centered on the y-axis. Then the equation of our parabola as a function of $x$ is given by $$y=d-\frac{x^2}{4d}$$

This is to be consistent with the information given about the parabola.

Next, let's consider the contribution that an infinitesimal amount of charge $dq=\lambda dl$ makes to the field using Coulomb's law, where $\lambda$ is the charge per unit length and $k=\frac{1}{4\pi \epsilon _0}$(or whatever units you prefer).

$$\mathrm{d}\vec E=\frac{k\lambda \mathrm{d}l}{r^2}\hat r=\frac{k\lambda (x\hat i+y\hat j) \mathrm{d}l}{(x^2+y^2)^{3/2}}$$

Now, by symmetry the x-component of our field is $0$, so we only need to focus on the y-component: $$\mathrm{d}\vec E=\frac{k\lambda y \mathrm{d}l}{(x^2+y^2)^{3/2}}\hat j$$

Next we need to get things in terms of just $x$ (which parametrizes our line to integrate over). From before we know that $y=d-\frac{x^2}{4d}$. We also need to integrate (add up) all of these contributions to the field along the curve of the parabola. To do this, we need to know what $\mathrm{dl}$ is in terms of $x$:

$$\mathrm{d}l=\mathrm{d}x\left[1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right]^{1/2}=\mathrm{d}x\left(1+\frac{x^2}{4d^2}\right)^{1/2}$$

Substituting this into the above expression for $d\vec E_y$ and integrating over the curve will give you the field at the origin.

• @Frobenius Is this better in terms of giving a solution? Aug 7, 2018 at 9:58
• @Frobenius Thanks! I have deleted the old version. Aug 7, 2018 at 11:12

The simplest way I can see is to use Coulomb's law $$\int_l\frac{\rho}{|\mathbf{r}|^2}\mathbf{r}\ d\mathbf{l}$$ with the curve parameterized by $\left(t,k-\dfrac{t^2}{4k}\right)$ for some $k$. (That parameterization guarantees the focus is the origin).

EDIT: (Thanks Aaron Stevens and tatan for the input)

Exploiting the symmetry of the problem shows the field will be in the vertical (in my calculation y)- direction, so we focus on the y-components of the contributions. $$dE_y=dE\sin\theta$$ $$\sin\theta=\frac{k-\frac{t^2}{4k}}{\frac{t^4}{16k^2}+\frac{t^2}{2}+k^2}$$ $$E_y=\frac{1}{4\pi\epsilon_0}\int\frac{dQ}{|\mathbf{r}|^2}\sin\theta$$ $$=\frac{1}{4\pi\epsilon_0}\int_{-\infty}^\infty \frac{k-\frac{t^2}{4k}\sqrt{1+\frac{x^2}{4k^2}}}{\left(\frac{t^4}{16k^2}+\frac{t^2}{2}+k^2\right)^{3/2}}\lambda\ dt$$

(We integrate all over the real line because the parabola is infinite).

I will leave that integral for you to do if you wish (it is pretty nasty but doable). The answer I get, which definitely should be checked, is $\boxed{\dfrac{\lambda}{3k\pi\epsilon_0}}$.

• You need to include the $\hat r$ vector here, or you could even use symmetry and only pick out the y-component of the field, since the field will point only in this direction. And you don't want your $dl$ to be a vector quantity. Aug 6, 2018 at 17:59
• Taking the limit of integration would definitely be something that would bother an inexperienced person here... so you may shed some light in that ;-) Aug 6, 2018 at 18:02
• I'll try attempting it again when I get the time, but for some reason I'm getting a divergent integral... could one of you post your solution?
– user195162
Aug 6, 2018 at 18:31
• It could diverge since you have an infinite amount of charge, but the line charge does go off to infinity, so it could also converge. I can't work through it at the moment, but I can later if no one else has replied. Aug 6, 2018 at 18:46
• Ok, so now the difference is in the differential length. The problem is that $dl$ and $dt$ are not the same. If you are defining parameter $t$ then $dl=\sqrt{(dx/dt)^2+(dy/dt)^2}dt$ Aug 6, 2018 at 23:23

Another alternative, that gives the same resuts, is to use parabolic cylindrical coordinates. In these coordinates, say $\tau, \sigma$ and $z$, we have the following relations to standard Cartesian coordinates:

\begin{eqnarray} x &=& \sigma \, \tau \, , \\ y &=& \frac{1}{2} \left(\tau^2 - \sigma^2\right) \, , \\ z &=& z \, , \end{eqnarray} with $\tau \in (-\infty,\infty), \, \sigma \in [0,\infty)$ and $z \in (-\infty,\infty)$. Fixing $\sigma = \sigma_0 \neq 0$ and $z = 0$, one finds curves given by \begin{equation} y = \frac{x^2}{2 \sigma_0^2} - \frac{\sigma_0^2}{2} \, , \end{equation} which are parabolas with their foci in the origin. In terms of the paramenters of the problem, $d = \frac{\sigma^2_0}{2}$ or $\sigma_0 = \sqrt{2d}$. Notice that both $\sigma$ and $\tau$ have units of Length$^{1/2}$. The corresponding scale factors are \begin{equation} h_\sigma = h_\tau = \sqrt{\tau^2 + \sigma^2} \, \, \, \mathrm{and} \, \, \, h_z = 1 \, . \end{equation} Thanks to this info, we can write the corresponding charge density \begin{equation} \rho(\mathbf{x}') = \lambda \, \frac{\delta(\sigma' - \sqrt{2d})}{\sqrt{\tau'^2 + \sigma'^2}} \, \delta(z') \, . \end{equation} Now, we can compute the electric field at the origin with \begin{equation} \mathbf{E}(\mathbf{x} = 0) = - \frac{1}{4 \pi \epsilon_0} \int \mathrm{d}^3 x' \, \rho(\mathbf{x}') \frac{\mathbf{x}'}{|\mathbf{x'}|^3} \, , \end{equation} and \begin{equation} \mathbf{x}' = \sigma' \tau' \, \mathbf{i} + \frac{1}{2} \left(\tau'^2 - \sigma'^2\right) \, \mathbf{j} + z' \, \mathbf{k} \, . \end{equation} Notice also that $\mathrm{d}^3 x' = (\tau'^2 + \sigma'^2) \, \mathrm{d} \tau' \, \mathrm{d} \sigma' \, \mathrm{d} z'$. Integration basically goes as explained in other answers, meaning that one can simply focus on the $\mathbf{j}$ component of $\mathbf{E}(\mathbf{x} = 0)$, as the others vanish.

EDIT BONUS:

As a bonus, you can use this info to go back to the magnetic field problem. For a uniform current $I$ along the parabola you can write the current density as \begin{equation} \mathbf{J}(\mathbf{x'}) = I \, \frac{\delta(\sigma' - \sqrt{2d})}{\sqrt{\tau'^2 + \sigma'^2}} \, \delta (z') \, \, \boldsymbol{\tau'} \end{equation} where we used the unit vector (see the wiki link at the beginning of the answer) \begin{equation} \boldsymbol{\tau'} = \frac{\sigma' \, \mathbf{i} \, + \, \tau' \, \mathbf{j}}{\sqrt{\tau'^2 + \sigma'^2}} \, . \end{equation} Then you can use this to compute \begin{equation} \mathbf{B}(\mathbf{x} = 0 ) = - \frac{\mu_0}{4 \pi} \int \mathrm{d}^3 x'\, \mathbf{J}(\mathbf{x'}) \times \frac{\mathbf{x'}}{|\mathbf{x'}|^3} \, . \end{equation}