Superposition of electric fields from uniform charge density

[jamdr]

Homework Statement

Ok, here's the problem. It deals with the superposition of electric fields from uniformly charged shapes: A uniformly charged infinite plane is located at z = 0, with a surface density of charge σ. A uniformly charged spherical shell with the same surface density is located at (0, 0, 3R), with radius R. Find the magnitude and direction of the electric field at the point (2R, 0, 3R). I've drawn a diagram:

Homework Equations

To find the electric field at the point, you just add the electric fields from the plane and the sphere:

[tex]E=\frac{\sigma}{2\epsilon_{0}} + \frac{Q}{4 \pi R^{2} \epsilon_{0}}[/tex]

The Attempt at a Solution

I'm confused about the direction, however, or how to finish solving this. I know I only need to count the z-component of the electric field from the infinite plane because the x and y cancel. But what about the sphere? Any help would be appreciated. Thanks!

 

[RoyalCat]

I'm pretty sure you can't treat the hemisphere as a point charge, I'll crunch the numbers in a sec, but I suggest you look at it yourself.

 

[jamdr]

I got the equation for the spherical shell from my textbook, which uses it as an example. It says this about it:

Electric field due to a uniformly charged spherical shell. Outside the shell, the field lines have spherical symmetry: they diverge from the origin. The field line pattern is the same as that due to a point charge at the origin, and the mathematical expression for the electric field is the same, too. Inside the shell, there are no field lines at all, and E=0.

Honestly, point charges are about as far as we've gotten in my class, so I think I should stick with that.

 

[jamdr]

Can I replace Q with this:

[tex]Q=\sigma 4\pi R^{2} [/tex]

That would mean the electric field from the spherical shell is:

[tex]E_{shell}= \frac{\sigma}{\epsilon_{0} } [/tex]

Although I'm still not sure about the direction part of if anything else I've done so far is correct.

 

[RoyalCat]

I got the equation for the spherical shell from my textbook, which uses it as an example. It says this about it:

Electric field due to a uniformly charged spherical shell. Outside the shell, the field lines have spherical symmetry: they diverge from the origin. The field line pattern is the same as that due to a point charge at the origin, and the mathematical expression for the electric field is the same, too. Inside the shell, there are no field lines at all, and E=0.

Honestly, point charges are about as far as we've gotten in my class, so I think I should stick with that.

Your textbook discusses a full hollow sphere. Here you're dealing with a hemisphere, which is a different matter altogether.

Or did you write something different and I misunderstood because of the diagram?

Right now I'm constructing a complicated integral to find the field distribution of this hemisphere, so I think that if need be, you should just say that it's a full hollow sphere because dealing with a hemisphere is outside the scope of your abilities.

If it is supposed to mean a full hollow sphere, then yes, treating it as a point charge is A-okay. :)

From there, there remains only the question of actually superposing the fields. Which means just adding the individual contributions together.

The uniformly charged plane only contributes in the z direction and the hollow sphere contributes along the line connecting its center with the point in question.

The reasoning for this is from Coloumb's Law for a point charge:

[tex]\vec F_e=\frac{KQq\hat r}{r^2}[/tex]

[tex]\vec E=\frac{KQ\hat r}{r^2}[/tex]

[tex]\vec E || \hat r[/tex]

Can I replace Q with this:

[tex]Q=\sigma 4\pi R^{2} [/tex]

That would mean the electric field from the spherical shell is:

[tex]E_{shell}= \frac{\sigma}{\epsilon_{0} } [/tex]

Although I'm still not sure about the direction part of if anything else I've done so far is correct.

No, that is not true.

The radius in the point-charge situation,
[tex]\frac{Q}{4\pi \epsilon_0 r^2}[/tex] is the distance from the point charge to the the point where you want to know the electrical field.

When calculating what Q actually is, you're referring to the surface area of the sphere, in which case, you need to use the radius of the sphere, R.

That's an important distinction to make, because it's a bad notation scheme that can get you mixed up in the middle of a test or exercise and burn off a lot of time.