# On The Potentials Of Conservative Fields

I was working on some E&M problems when I came across one I did not quite understand, the problem goes something like this:

The potential at a radial distance r from a point charge q is V=$$\frac{q}{4\pi\epsilon_0r}$$ . Considering the absence of radial symmetry find the vector form of the electric field (E) from this expression for V

Usually, this is how I would have started this:

->first i know that E= -$$\nabla V$$ and the general property that $$\nabla (r^n)=nr^{n-1}\hat{r}$$

-> expressing V=$$\frac{q}{4\pi\epsilon_0r}$$ as V=$$\frac{q}{4\pi\epsilon_0}*r^{-1}$$, I get that

E= -$$\nabla V$$=$$\frac{q}{4\pi\epsilon_0r^2}\hat{r}$$

which is my expression for the vector field, E. However, I am not so sure about the following points:

(a) Over here, as we have considered a point charge, we are dealing with spherical/radial symmetry right? that's why we can apply Gauss's Law and get E= -$$\nabla V$$=$$\frac{q}{4\pi\epsilon_0r^2}\hat{r}$$ in the radial direction even without starting from the potential expression.

(b) What happens to my expressions when I do consider the absence of radial symmetry as the statement of the problem has mentioned? I tried looking it up online but to no avail. I just can't understand what is being asked from me; if I do neglect radial symmetry then my potential expression won't be the same right?

(c) Maybe I did not catch it, but are radial and spherical symmetry completely different things? here I thought in the context of Gauss's Law it would be the same.

• I don't quite understand what you mean by "What happens to my expressions when I do consider the absence of radial symmetry as the statement of the problem has mentioned?" Oct 20, 2020 at 14:35
• like normally the expression for E i got was assuming spherical symmetry, because it is a point harge and all. But, if i leave out that form of symmetry what do i get?
– F.N.
Oct 21, 2020 at 12:30

(a) Over here, as we have considered a point charge, we are dealing with spherical/radial symmetry right? that's why we can apply Gauss's Law and get $$E= -∇V=\frac{q}{ 4πϵ_0r^2}$$ in the radial direction even without starting from the potential expression.

1. Depending on the coordinate system, the operations you do for gradients are different but the important thing is that the gradient is the same no matter what coordinate system you evaluate it in given that you use the correct formula for evaluating it in that particular coordinate system. In the particular example that you've given in the post, it's more natural to use spherical coordinates.

Illustration:

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

Since our potential is only dependent on $$r$$, we only have to consider the first derivative operation in the expression abvoe:

$$\nabla V= (\frac{ \partial }{\partial r} \hat{r} + \frac{1}{r} \frac{ \partial }{\partial \theta} \hat{\theta} + \frac{1}{r \sin \theta} \frac{ \partial }{\partial \theta} \hat{\phi})V=-\frac{1}{4 \pi \epsilon_o} \frac{q}{r^2} \hat{r}$$

But suppose I evaluated this using gradient operator in Cartesian Co-ordinates:

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

First we'd have to expression the potential in cartesian coordinates as follows:

$$V =\frac{q}{4 \pi \epsilon_o r}=\frac{q}{4 \pi \epsilon_o r} \frac{q}{ \sqrt{x^2 +y^2 + z^2} }$$

Now applying the gradient in cartesian:

$$\nabla V = - \frac{ q}{ 4 \pi \epsilon_o} [ \frac{x \vec{i} }{(x^2 +y^2 +z^2)^{\frac32} } + \frac{y \vec{j}}{(x^2 +y^2 +z^2)^{\frac32} } + \frac{z \vec{k}}{(x^2 +y^2 +z^2)^{\frac32} }]$$

Now recall:

$$\hat{r} = \frac{ x \vec{i} + y \vec{j} + y \vec{k} } { \sqrt{x^2 +y^2 + z^2 } }$$

Hence,

$$\nabla V = \frac{q}{4 \pi \epsilon_o r} \hat{r} \checkmark$$

The derivative in both coordinate systems is the same! Hence the gradient operation is one that is invariant of coordinate systems. You only have to make sure that you use the correct recipe.

(b) What happens to my expressions when I do consider the absence of radial symmetry as the statement of the problem has mentioned? I tried looking it up online but to no avail. I just can't understand what is being asked from me; if I do neglect radial symmetry then my potential expression won't be the same right?

If you wish to evaluate the gradient in a different coordinate system, you are free to do so but be careful to use the correct recipe!

(c) Maybe I did not catch it, but are radial and spherical symmetry completely different things? here I thought in the context of Gauss's Law it would be the same.

Radial means that you can rotate the configuration about a point in a plane and still the electric field would be invariant. Usually associated with cylinders; you spin the cylinder about its axis and it looks the same

Spherical symmetry usually means you can rotate the configuration in 3-d space about a point and the electric field would be invariant. Usually associated with spheres; you spin the sphere around the axis passing through its center and look the same.

Note: Usually, spherical symmetry implies cylindrical symmetry but cylindrical symmetry doesn't imply spherical symmetry.

I found more discussion about symmetries here

For a more thorough presentation of the ideas I said above, See Griffith's introduction to electrodynamics

A comment:

Think of it like this: the electric field is a real meaningful quantity hence it must be the same no matter it must not change depending on which coordinate system which you take derivative in.

• So based on this and the previous answer, i just need to use the gradient operator in different coordinates right? because i am not using radial symmetry, i just turn to cartesian and show that i get the same answer. The other parts i understand finally. About Griffiths, yea thanks for the heads up, definitely going to check it[i actually have a copy nearby, although i am not taking that class until a bit later down the line]
– F.N.
Oct 19, 2020 at 8:30
• Yes, you really are not applying any sort of mathematical symmetry other than the fact that the gradient is easier to evaluate in the spherical coordinate system in this case. Oct 19, 2020 at 8:32
• great, i will just finish this problem right away then. So it all boil downs to using $\nabla$ in different systems based on what i need.. nice!
– F.N.
Oct 19, 2020 at 8:34
• Actually, this concept is very well discussed in a book called Pavel Grinfield : An introduction to tensor analysis. The book starts with the motivation behind finding a way to create a formula that spits out the correct operations for gradient no matter what coordinate system you use. So, if you're interested in exploring that idea more I suggest you check that book. Oct 19, 2020 at 8:37
• I will definitely check it out later, thanks for the input.
– F.N.
Oct 19, 2020 at 13:25

(a) Over here, as we have considered a point charge, we are dealing with spherical/radial symmetry right? thats why we are able to apply Gauss's Law and get $$E= -\nabla V= \frac{q}{4π \epsilon_0 r^2}\hat r$$ in the radial direction

That is correct.

(b) What happens to my expressions when i do consider absence of radial symmetry as the statement of the problem has mentioned? I tried looking it up online but to no avail. I just can't understand what is being asked from me; if i do neglect radial symmetry then my potential expression won't be the same right?

This would be the case if you considered a non-point source charge with varying charge density across it. In this case you would need to express the electric field in terms of the two other components $$\hat \theta$$ and $$\hat \phi$$.

(c) Maybe i did not catch it, but are radial and spherical symmetry completely different things? here i thought in the context of Gauss's Law it would be the same.

In the context of this problem (point charge), radial and spherical symmetry are the same. If you considered the value of the electric field at any constant $$r$$ the value of the field will be the same.

• I See. In regards to (b), this means i just need to use the $\nabla$ operator,expressed in other coordinate system(based on $\phi and \theta$, on the expression for potential right? because i am neglecting radial symmetry while still using a point charge, so i just need to show i get the same expression for E anyway correct?
– F.N.
Oct 19, 2020 at 8:23
• That is correct. Cheers. Oct 19, 2020 at 8:33
• cheers mate!🥃🥃
– F.N.
Oct 19, 2020 at 8:35