Spherical coordinates choice for an electric field problem

In summary, I am finding the electric field from a spherical shell at a point on the z-axis outside the shell. The shell is centered at the origin, and I am only allowed to use coulomb's law. I want to find dE in spherical coordinates first then transform it to Cartesian before integrating to get E. To find dE in spherical coordinates, I use R'_z = R'_r cos(Θ) - R'_Θ sin(Θ). I get R'_z = (z-R)cos(Θ) + Θsin(Θ), which is incorrect. The correct expression is R'_z =
  • #1
user3
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I am finding the electric field from a spherical shell at a point on the z-axis outside the shell. The shell is centered at the origin,and I am only allowed to use coulomb's law. I want to find dE in spherical coordinates first then transform it to Cartesian before integrating to get E.

So I choose the spherical coordinates of the point at which I want to find E to be (z , 0 , ø), and the coordinates of a piece of charge on the surface of the shell to be (R, Θ , ø), where R is the radius of the shell and z is the distance on the z axis above the origin.

When I transform to Cartesian coordinates to get E_z, I use R'_z = R'_r cos(Θ) - R'_Θ sin(Θ). I get
R'_z = (z-R)cos(Θ) + Θsin(Θ), Which is incorrect. The correct expression is R'_z = z-Rcos(Θ) .

R' is the distance vector from the source to the point.

I believe I am doing something wrong with choosing the coordinates, but I can't really tell where. Can somebody help me ?
 
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  • #2
The result for R'z should follow easily from the figure.
 

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  • #4
OK. I don't think I understand what you are trying to do.

It would probably help if you can show the steps that you used to get your incorrect result. Is there a particular reason why you want to do it this way?
 
  • #5
I know the final step is incorrect because it's not Z-Rcos(Θ), I can't understand where my sequence fails though. I want to do it this way just to practice working in spherical coordinates
 

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  • #6
Consider the vector ##\small \vec{R}## that locates dQ on the surface of the sphere. The vector points from the origin to a point that has spherical coordinates ##\small (R, \theta, \phi)##. That doesn't mean that the vector ##\small \vec{R}## can be expressed as ##\small \vec{R} = R\hat{r}+\theta \hat{\theta} + \phi \hat{\phi}##. If ##\small \hat{r}## is the radial unit vector at the location of dQ, then you could write ##\small \vec{R} = R\hat{r}##.

If you have a vector ##\small \vec{A}## that points from the origin to a point with spherical coordinates ##\small (r_1, \theta_1, \phi_1)## and another vector ##\small \vec{B}## that points from the origin to spherical coordinates ##\small (r_2, \theta_2, \phi_2)##, you can't subtract the vectors by subtracting the spherical coordinates.

##\vec{A} - \vec{B} \neq (r_1-r_2) \hat{r}+(\theta_1 - \theta_2) \hat{\theta} + (\phi_1 - \phi_2) \hat{\phi}##

##\vec{A} - \vec{B}## is not represented by ##(r_1-r_2, \theta_1 - \theta_2, \phi_1 - \phi_2)##
 
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  • #7
If you mean that we have to transform the two spherical coordinate points to Cartesian before we get the distance vector between them, then what good is it to have two points given in spherical coordinates ?

Also, what is the reference of the vector (R, Θ, ø)?

ie: (X,Y,Z) = (X,Y,Z)-(0,0,0) = X a_x + Y a_y + Z a_z

in Cartesian coordinates , but is

(R, Θ, ø) = (R, Θ, ø)-(0,0,0) = R a_r + Θ a_Θ + ø a_ø in spherical coordinates ?

if yes, then why can't we substract some other vector from (R, Θ, ø) instead of (0,0,0)?
 
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  • #8
user3 said:
If you mean that we have to transform the two spherical coordinate points to Cartesian before we get the distance vector between them, then what good is it to have two points given in spherical coordinates ?

For getting the distance between two points, spherical coordinates are not convenient. But spherical coordinates are useful for many other things.

Also, what is the reference of the vector (R, Θ, ø)?

ie: (X,Y,Z) = (X,Y,Z)-(0,0,0) = X a_x + Y a_y + Z a_z

in Cartesian coordinates , but is

(R, Θ, ø) = (R, Θ, ø)-(0,0,0) = R a_r + Θ a_Θ + ø a_ø in spherical coordinates ?

No. By constructing your own examples, you should be able to see that this doesn't make sense.
Any vector ##\vec{A}## that points from the origin to any point, p, is "radial". That is, it points "away from the origin". If you construct the unit vectors ##\hat{a}_r, \hat{a}_\theta, \hat{a}_\phi## at the point p, then the vector would be ##\vec{A} = A\hat{a}_r##, where ##A## is the magnitude of the vector. This would be true no matter what the values of (r, Θ, ø) for the point p.
 
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  • #9
Thank You for your time
 

What are spherical coordinates?

Spherical coordinates are a system of locating points in three-dimensional space using two angles and a distance from a fixed point, also known as the origin. The angles are typically referred to as azimuth and elevation, and the distance is known as the radius.

When should spherical coordinates be used for an electric field problem?

Spherical coordinates are useful for problems that involve spherical symmetry, such as a point charge or a charged sphere. They can also be used when the electric field is being measured at a specific distance from the origin.

How do you convert from Cartesian coordinates to spherical coordinates?

To convert from Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, ϕ), you can use the following equations:
r = √(x² + y² + z²)
θ = arccos(z/r)
ϕ = arctan(y/x)

What are the advantages of using spherical coordinates for an electric field problem?

Spherical coordinates allow for easier visualization of the electric field, as it is often easier to imagine a spherical shape rather than a Cartesian grid. They also simplify calculations for problems with spherical symmetry, making them more efficient to solve.

Are there any limitations to using spherical coordinates for an electric field problem?

Spherical coordinates may not be the best choice for problems that do not have spherical symmetry, as the calculations can become more complicated. They are also not as commonly used as Cartesian coordinates, so it may be more difficult to find resources or support for solving problems using spherical coordinates.

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