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This question already has an answer here:

What is the $\hat{r}$ (vector) in the formula $\vec{E} = k\frac{q}{r^2} \hat{r}$ for the electric field ? Why we dont use the vectors $\vec{i},\ \vec{j},\ \vec{k}$? Also why this vector doesn't appear in the formula where field is the gradient of potential where we have $\vec{i},\ \vec{j},\ \vec{k}$ (unit vectors)?

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marked as duplicate by Buzz, Jon Custer, Kyle Kanos, ZeroTheHero, stafusa Feb 2 at 0:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Which formula are you referring to and where did you find it? $\endgroup$ – my2cts Jan 31 at 17:27
  • $\begingroup$ softschools.com/formulas/images/electric_field_formula_5.png $\endgroup$ – ado sar Jan 31 at 17:53
  • $\begingroup$ I still don't see a formula in your question. $\endgroup$ – my2cts Jan 31 at 18:06
  • $\begingroup$ I think he’s talking about $\nabla=\hat{i}\frac{\partial}{\partial x}...$. $\endgroup$ – G. Smith Jan 31 at 18:12
  • $\begingroup$ @adosar, you can write the gradient operator in spherical coordinates and then it involves the spherical unit vectors. $\endgroup$ – G. Smith Jan 31 at 18:13
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You can write the field of a point charge $q$ (at the origin) in Cartesian coordinates as

$$\mathbf{E}=q\frac{x\hat{i}+y\hat{j}+z\hat{k}}{(x^2+y^2+z^2)^{3/2}}$$

or you can use spherical coordinates to write it as

$$\mathbf{E}=\frac{q}{r^2}\hat{r}.$$

The latter is simpler and easier to remember, and it makes the spherical symmetry obvious.

(By the way, I’ve written these formulas in Gaussian units, where the Coulomb constant is 1. Stick a $k$ or a $\frac{1}{4\pi\epsilon_0}$ in front if you prefer other unit systems.)

As ZeroTheHero explained, $\hat{r}$ is a radial unit vector. In spherical coordinates there are also tangential unit vectors $\hat{\theta}$ and $\hat{\phi}$, but you don’t need these to write a purely radial field, such as for a point charge.

If you are deriving the field from the potential using $\mathbf{E}=\vec\nabla\varphi$, you can write the gradient operator in Cartesian coordinates as

$$\vec\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial x}$$

or in spherical coordinates as

$$\vec\nabla=\hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}+\hat{\phi}\frac{1}{r\sin{\theta}}\frac{\partial}{\partial \phi}.$$

https://en.wikipedia.org/wiki/Spherical_coordinate_system

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In many E&M textbooks:

  1. One tends not to use $i$ because it's also used for current.
  2. One tends not to use $j$ because it's also used for current density.
  3. One tends not to use $k$ because it's also used for wave vectors.

$\hat r$ is the unit vector in the radial direction in spherical coordinates.

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    $\begingroup$ In Cartesian coordinates, $\hat{i}$, $\hat{j}$, and $\hat{k}$ are sometimes used as unit vectors, rather than $\hat{x}$, $\hat{y}$, and $\hat{z}$. $\endgroup$ – G. Smith Jan 31 at 17:46
  • $\begingroup$ @G.Smith, I think it's clear that Zero knows thta, but is explaining why it would be confusing to use those names in an electromagnetics problem. $\endgroup$ – The Photon Jan 31 at 18:14
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    $\begingroup$ Using $i$, $j$, and $k$ with hats to make them unit vectors is no more confusing than using $x$, $y$, and $z$ with hats to make them unit vectors. In both cases, if you leave off the hat you get something else. $\endgroup$ – G. Smith Jan 31 at 18:32
  • $\begingroup$ @G.Smith IMO the core problem is with math people using $\hat i$, $\hat j$ and $\hat k$, not with physics people using $\hat x$, $\hat y$ and $\hat z$, which is a lot more natural: after all we talk of the $x$-axis and the $x$-component. Anyways... In E&M it's better to stay away from $i$, $j$ $k$ because of other usage, like $j$ is also the imaginary unit because $i$ is already the current. We can agree that notation of unit vector is the least of problems in E&M. $\endgroup$ – ZeroTheHero Jan 31 at 18:51
  • $\begingroup$ @ZeroTheHero I'm not sure that I see your reasoning here. After all, $z$ and $x$ are used as symbols for impedance and reactance, respectively, in AC circuits, and $y$ is used to refer to rapidity in relativistic electrodynamics. Ultimately, it's probably an arbitrary convention. $\endgroup$ – probably_someone Jan 31 at 19:05

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