Electromagnetism: cylindrical or rectangular coordinates.

[Bromio]

Hi.

Sorry my spelling, because I am not English.

Homework Statement

In a sphere truncated sector with an angle of 60 degrees, there is a uniform charge distribution, $$\rho$$. Calculate the electric field in $$(0,0,0)$$. The sector starts in $$z=a$$ and ends in $$z = b$$. The sphere center is in $$(0,0,0)$$.

Homework Equations

To calculate the electric field, I use:
$$\bar{E}(\bar{R}) = \displaystyle\frac{1}{4 \pi \epsilon_0} \displaystyle\int\displaystyle\int\displaystyle\int_{V'} \rho\left(\bar{R}\right) \displaystyle\frac{\bar{R}-\bar{R}'}{\left|\bar{R}-\bar{R}'\right|^3}\;dV'$$

The Attempt at a Solution

I know that:
$$\rho\left(\bar{R}\right) = \rho$$

$$\bar{R} = 0$$

$$\bar{R}' = \hat{R}'R'$$

$$dV' = R'^2\sin\theta\;dRd\theta d\phi$$

Now, I substitute these values in the integral and I solve it.

What is the problem? I don't know if I can integrate using cylindrical coordinates directly or, on the other hand, if I must use rectangular coordinates:
$$\bar{E}(0) = \displaystyle\frac{1}{4 \pi \epsilon_0} \displaystyle\int_0^{2 \pi}\displaystyle\int_0^{\pi /6}\displaystyle\int_a^b \rho\displaystyle\frac{\hat{R}R'}{R'^3}R'^2 \sin \theta\;dRd\theta d\phi$$

or

$$\bar{E}(0) = \displaystyle\frac{1}{4 \pi \epsilon_0} \displaystyle\int_0^{2 \pi}\displaystyle\int_0^{\pi /6}\displaystyle\int_a^b \rho\displaystyle\frac{-\hat{z}R' \cos \theta}{R'^3}R'^2 \sin \theta\;dRd\theta d\phi$$ ?

In first situation, the result will depend on $$\hat{R}$$, but, in the second one, it will depend on $$\hat{z}$$. Which is the correct solution and why?

Thank you.
Bromio.

 

[vela]

Your post is a bit unclear. You refer to cylindrical coordinates, but your integrals are written using spherical coordinates.

Homework Statement

In a sphere truncated sector with an angle of 60 degrees, there is a uniform charge distribution, $$\rho$$. Calculate the electric field in $$(0,0,0)$$. The sector starts in $$z=a$$ and ends in $$z = b$$. The sphere center is in $$(0,0,0)$$.

I'm not sure I understand what region the charge occupies. You have these limits z=a and z=b, yet later on, you use these values as the limits for the radial coordinate r, not z.

Homework Equations

To calculate the electric field, I use:
$$\bar{E}(\bar{R}) = \displaystyle\frac{1}{4 \pi \epsilon_0} \displaystyle\int\displaystyle\int\displaystyle\int_{V'} \rho\left(\bar{R}\right) \displaystyle\frac{\bar{R}-\bar{R}'}{\left|\bar{R}-\bar{R}'\right|^3}\;dV'$$

The Attempt at a Solution

I know that:
$$\rho\left(\bar{R}\right) = \rho$$

$$\bar{R} = 0$$

$$\bar{R}' = \hat{R}'R'$$

$$dV' = R'^2\sin\theta\;dRd\theta d\phi$$

You have several problems from confusing R and R'. In the integral, the charge density should be a function of R', not R. In the volume element dV', you have both R and R'. It should just be R', right?

 

[Bromio]

Thanks for answering, vela.

Some of the errors you have detected are misprints, so I'll correct them. In addition, I thought in spherical coordinates, but I wrote cylindrical ones. Sorry, again. Now I attach an image in order to see how is the situation.

$$\rho\left(\bar{R}'\right) = \rho$$
$$dV' = R'^2\sin\theta\;dR'd\theta d\phi$$The question is the same I asked above.

Thank you.
Bromio

 

[vela]

In your first integral, the unit vector should be $$\hat{R}'$$, not $$\hat{R}$$. I assume that's just another typo as you wrote it correctly earlier. You also dropped an overall negative sign.

Other than those errors, both expressions are correct. In the first integral, you're not taking advantage of the axial symmetry of the problem yet. If you plug in the expression for $$\hat{R}'$$ and integrate, you'll find the x and y components come out to be zero. You can, of course, prior to integration, argue those components should be 0 by symmetry and just calculate the z component. That's what the second integral gives you.

 

[Bromio]

Thank you, vela

You are right: the error you indicates is a misprint again (in paper I've written the equation correctly).

Because of your answer, I understand I should substitute $$\hat{R}'$$ unit vector by $$\hat{x} \cos \theta \sin \phi + \hat{y} \cos \theta \cos \phi + \hat{z} \sin \theta$$ and then, integrate. Why I must do it?

In other words:
Is this solution correct?

$$\bar{E}(0) = \displaystyle\frac{1}{4 \pi \epsilon_0} \displaystyle\int_0^{2 \pi}\displaystyle\int_0^{\pi /6}\displaystyle\int_a^b \rho\displaystyle\frac{\hat{R}\cancel{R}}{\cancel{R^3}}\cancel{R^2} \sin \theta\;dRd\theta d\phi$$

Or should I write $$\hat{R}$$ in rectangular coordinates?

Thank you.
Bromio

PS: I don't know what happens with LaTeX, but some formulas doesn't appear properly, although I have checked them in an auxiliary program.

 

[vela]

You're still missing the overall negative sign, and the unit vector you should be

$$\hat{R} = \sin\theta\cos\phi\,\hat{x} + \sin\theta\sin\phi\,\hat{y} + \cos\theta\,\hat{z}$$

I'm not sure about your question about why you would do this. How else are you going to evaluate the integral? Note that the unit vector depends on the variables of integration, so you can't just pull it out of the integral.

 

[Bromio]

Thank you, vela

The integral is $$\bar{E}(0) = \displaystyle\frac{1}{4 \pi \epsilon_0} \displaystyle\int_0^{2 \pi}\displaystyle\int_0^{\pi /6}\displaystyle\int_a^b \rho\displaystyle\frac{-\hat{R}R'}{R^3}R'^2 \sin \theta\;dRd\theta d\phi$$

I'm not sure about your question about why you would do this. How else are you going to evaluate the integral? Note that the unit vector depends on the variables of integration, so you can't just pull it out of the integral.

That's my trouble! So, I can't pull $$\hat{R}$$ out because of its dependency.

For example, if I were calculating the electric field in a charged sphere, obvioulsy the field depends on radial direction and there is spherical simmetry, so I could integrate the $$\hat{R}$$ unit vector. Yes?

Thanks, again.
Bromio.

 

[vela]

That's my trouble! So, I can't pull $$\hat{R}$$ out because of its dependency.

For example, if I were calculating the electric field in a charged sphere, obvioulsy the field depends on radial direction and there is spherical simmetry, so I could integrate the $$\hat{R}$$ unit vector. Yes?

I'm not sure what you're trying to ask here.

 

[Bromio]

I'm not sure what you're trying to ask here.

I've considered a different problem. Suppose a sphere with a volumetric charge distribution, $$\rho$$. I want to calculate the field in every point of the space. Generally, I solve this problem using Gauss' law, but suppose I want to use electric field formula. Obviously, $$\bar{E}$$ direction will be $$\hat{R}$$, radial. Could I pull $$\hat{R}$$ out the integral? Or should I substitute it for its rectangular coordinates?

Thank you.
Bromio.

 

[vela]

Your integral in post 7 has mistakes. I assumed you were just being careless, but your subsequent questions make me question my assumption. R and R' aren't interchangeable, and I think you're confusing yourself by not being keeping track of which is being used where.

There's nothing particularly special about the unit vector. Just like any other quantity in the integrand, if it depends on the variables of integration, you can't pull it out of the integral. It's that simple. If you want to evaluate the integral, you'll need to express the unit vector explicitly in terms of the integration variable.

 

[Bromio]

Your integral in post 7 has mistakes. I assumed you were just being careless, but your subsequent questions make me question my assumption. R and R' aren't interchangeable, and I think you're confusing yourself by not being keeping track of which is being used where.

I think $$\bar{R}'$$ points at the source and $$\bar{R}$$ points to where I want to know the electric field.

So, the integral should be:
$$\bar{E}(0) = \displaystyle\frac{1}{4 \pi \epsilon_0} \displaystyle\int_0^{2 \pi}\displaystyle\int_0^{\pi /6}\displaystyle\int_a^b \rho\displaystyle\frac{-\hat{R}'R'}{R'^3}R'^2 \sin \theta\;dR'd\theta d\phi$$

In this case, $$\bar{R} = 0$$. Is it correct?

There's nothing particularly special about the unit vector. Just like any other quantity in the integrand, if it depends on the variables of integration, you can't pull it out of the integral. It's that simple. If you want to evaluate the integral, you'll need to express the unit vector explicitly in terms of the integration variable.

Thank you. I've understood this finally.

Thank you.
Bromio.