Electric force between two halves of a sphere

[darkSun]

Homework Statement

(I'll be using k=1/4pi*permittivity of free space)
Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. Express the answer in terms of the radius R and the total charge Q.

I tried to show how I attempted the problem, but trying to follow that, especially written like this, may be inconvenient, so if someone could lead me in the right direction that'd cool. The way I tried it may not be correct anyway.

Homework Equations

This equation is from another problem, it looked useful so:
the force per unit area on a charged slab thickness a is:
integral from 0 to a of rho times the electric field times dx
rho is the volume charge density

The Attempt at a Solution

What I tried to do is make a cylindrical column from the north pole to the center. I then used the above formula, and got the force per unit area to be 3kQ^2 / 8piR^4. To get this I integrated Q over the volume of a sphere, which is rho, times the formula for the electric field of a uniformly charged sphere inside the sphere, kQr/R^3. Then, I multiplied by a surface area element in spherical coordinates (with phi as azimuthal angle), r^2 sin phi dphi dtheta. Then I multiplied by cos phi, because I want the upward force component. I integrated this over the top half of the sphere (phi goes from 0 to pi/2, and theta from 0 to 2pi).

I obtained 3kQ^2/8R^2, while the book says the right answer is half that!
It's from Griffith's Intro to Electrodynamics, problem 2.44 if that helps.

 

[tiny-tim]

Hi darkSun!

(have a pi: π and try using the X² tag just above the Reply box )

Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. Express the answer in terms of the radius R and the total charge Q.
…
the force per unit area on a charged slab thickness a is:
integral from 0 to a of rho times the electric field times dx
rho is the volume charge density
…
What I tried to do is make a cylindrical column from the north pole to the center. …

I don't think there's any convenient short-cut for this problem …

I think you'll just have to integrate the (component of) force from each point on one sphere to each point on the other sphere.

 

[darkSun]

Okay. I tried doing that but it seems as though it gets too complicated. It's hard to integrate the resulting expression in rectangular coordinates, and in spherical I can't even get an expression for the points from one hemisphere to another. I didn't think the problem was this crazy lol. Thanks for your help tiny-tim.

 

[gabbagabbahey]

I'm not sure if you are still working on this problem, but in case you are, you were very close with your first attempt. There is nothing wrong with using the electric field due to the entire sphere $\textbf{E}_{total}$ (as opposed to the field due to just the southern hemisphere $\textbf{E}_{S}$) because when you integrate over the northern hemisphere the part of the field from that hemisphere ($\textbf{E}_{N}$) will not exert a force on it.

In other words, you want to calculate

$$\int_{\text{N-Hemi}} \rho\textbf{E}_{S}d\tau$$

but since an object can't exert a force on itself, you know that

$$\int_{\text{N-Hemi}} \rho\textbf{E}_{N}d\tau=0$$

and so there is no harm in adding that to the integral you want to calculate:

$$\textbf{F}=\int_{\text{N-Hemi}} \rho\textbf{E}_{S}d\tau=\int_{\text{N-Hemi}} \rho\textbf{E}_{S}d\tau+\int_{\text{N-Hemi}} \rho\textbf{E}_{N}d\tau=\int_{\text{N-Hemi}} \rho(\textbf{E}_{S}+\textbf{E}_{N})d\tau=\int_{\text{N-Hemi}} \rho\textbf{E}_{total}d\tau$$

The problem with your first attempt was that you were only integrating over the outer surface of the northern hemisphere instead of integrating over the volume of the hemisphere (the southern hemi will exert a force on every piece of the hemisphere, not just the pieces on the outer surface).

 

[darkSun]

Yes, this problem has still been bothering me. Hmm, when you phrase it like that, gabbagabbahey, it seems so obvious! But it's still a useful thing to realize.

Anyway, I did the integral the way you said to, and I got 3Q^2 / 64*pi*epsilon*R^2. That is the book's answer with an extra 4 in the denominator. T_T

I also don't think my first attempt was watertight, but what I was trying to do was not to figure out force per area, but force per cylinder going through the center, and adding up the cylinders. Eh, if it's not right, it's not right.

 

[tiny-tim]

… an object can't exert a force on itself …

hey, that's neat!

 

[mathfeel]

There is a trick to this problem.

First the electric field of a charge ball of radius 1 is:
$$E(r)= \begin{cases} \dfrac{k Q}{r^2} & r > 1 \\ k Q r & r <1 \end{cases}$$

Suppose you cut a sphere along the x-y plane into two hemispheres then displace them by a small distance dz.

This is do not actually change the electric field much, except now one will have some additional electric field on the x-y plane. The energy stored in the additional field is how much work done to pull them apart by dz.

The force is of course just dW/dz.

$$\begin{aligned} dW = \frac{1}{8\pi k} \int E^2 dV & = \frac{1}{8\pi k} \int E^2(r) 2\pi r dr dz \\ & = \frac{k Q^2}{4} \left( \int_0^1 r^3 dr + \int_{1}^{\infty} r^{-3} dr \right) dz \\ F_z = \frac{dW}{dz} &= \frac{k Q^2}{4} \left( \frac{1}{4} + \frac{1}{2} \right) =\frac{3 k Q^2}{16} \end{aligned}$$

By the way, I think the "supposed way" to do this Griffith problem is to use the formula for surface pressure of a conductor earlier in the chapter and add up the integrate z-component of the force on the base and the side of a hemisphere. Be ware, you need to make a correct physical argument about the direction of the pressure/force.