Electrostatics: a question regarding a charged insulated sphere

[roya]

the question that i have encountered was:
"given an insulated charged sphere, what is the force applied by one half of the sphere on the other half?"

i have tried calculating the integral but it seemed a little too complicated for me (plus i am not absolutely confident what the correct integration actually is)
is there a better way approaching this question?

 

[granpa]

what is the force on each individual charge dq?

 

[roya]

the question is what is the entire force "sensed" by half a sphere
the only parameter given is the charge density p (forgot to mention)

 

[granpa]

imagine the sphere is filled with fluid under pressure. that should give the same result for the force at the surface (where the charge is). it should be easy to calculate the force on a plane down the center.

you understand that all the charge is evenly distributed at the surface right?

 

[roya]

why would the charge be distributed at the surface if the sphere is insulated?

 

[granpa]

its easy to see if the sphere is a conductor. there can't be any electric field within the conductor. I don't think it makes any difference its an insulator.

what class is this for?

 

[granpa]

wait. charge density? so its asking about a hypothetical sphere with uniform charge density throughout? well that would be different. I'm not sure how to solve that.

 

[roya]

electromagnetism

if the sphere was a conductor then i would have used the pressure sensed by each half a sphere in order to calculate the force (as you suggested), but since the sphere is insulated i can't seem to come up with an idea on how to solve this ...

 

[roya]

insulated sphere with a uniform charge density p, correct.
thanks for your help, i hope someone else has an idea though.

 

[granpa]

insulated? I don't see what that has to do with anything. it might help if you just type in the problem as it is stated in the book.

 

[granpa]

I don't see any easy solution but you can always just use the pressure technique to calculate the force from each infinitesimally thick shell and integrate over all r. maybe someone can come up with something better though.

 

[roya]

the fact that it is insulated and not conducting makes it possible for the charge to be uniformly distributed throughout the entire sphere. if it was not insulated then charges would not have been bounded and there would have been no electric field inside.
the question is originally stated in a different language, i merely translated it into english.
anyway, thanks again.

 

[granpa]

well I don't know what 'insulated' means in your language but in English it means that the sphere is surrounded by an insulating material. and even if it were itself an insulating material the charge would ordinarily still be at the surface.

 

[granpa]

would it help if you reverse it and think in terms of gravity and pressure as a function of depth?

 

[roya]

i am not completely sure how that helps... maybe if you could be more specific?

 

[granpa]

the force of attraction of the 2 halves must equal the pressure across a plane through the center. simply reverse everything and you have your original problem. attraction becomes repulsion, etc...

 

[Defennder]

Do you have Mathematica or something? You could do it in Cartesian coordinates, or spherical coordinates, but since you're dealing with two hemispheres, I'll prefer Cartesian. But then again I'm not familiar with volume integration using spherical basis vectors since \mathbf{a_{\phi}} and the other basis vectors are not constant for varying values of \phi and \theta.

If you got this question from a textbook, there's a very good chance that you can exploit some law or theory to cut down on a lot of integration and vector calculus.

But otherwise I'll start with \mathbf{E}(x,y,z) = \int_{vol} \frac{\rho}{4\pi \varepsilon} \frac{\mathbf{r}-\mathbf{r'} }{|\mathbf{r}-\mathbf{r'}|^3} dv'. Limits of the integration is confined to the left half of the hemisphere, so y,z ranges from -r to r, x from -r to 0. Once you get the field (by Mathematica or by tedious hand calculations), you'll need to integrate for the force F using d\mathbf{F} = p\mathbf{E}dv which is a volume integration over the other half of the hemisphere.

As said before, this is really tedious without Mathematica or some symmetrical considerations.

[quote="granpa]the force of attraction of the 2 halves must equal the pressure across a plane through the center. simply reverse everything and you have your original problem. [/quote]Hi, I'm not sure what force of attraction you're talking about. The charge is uniform throughout without any change in sign, is it not?

 

[granpa]

see post 14

 

[granpa]

the electric field is simply proportional to r (up to the surface) and always directed outward. just like gravity.

and the pressure below ground is simply equal to the weight of material in a column (not wedge) directly above it.

 

[roya]

so a plane in between the two halves will experience pressure due to the force of repulsion, that makes sense, but i can't figure out how to calculate the pressure without finding the electric field (or force for that matter)
if you meant something else then i probably just didn't get it

 

[granpa]

in a hollow sphere filledwith pressurized fluid the force on the surface is the same as the force acting on the charged particles on th e surface of your sphere. the net force from one hemispher on the other must be the same therefore as the forec of pressure across a plane dividing them. there is no pressure in electrostatics. its simply a fact that the 2 forces must be equal.

reversing it we can imagine that the surface charges are masses attractang one antother and the pressure within is produced that way.

 

[granpa]

what is the force of attraction between the 2 hemispheres of the earth? it obviously must equal the force due to pressure with which they repel one another. calculating the pressure as a function of depth should be easy

 

[weejee]

Calculate the E field and perform a volume integration of the (E field)x(charge densitity) upon an entire hemisphere.

The point is that you don't have to try to calculate the effect of the other hemisphere only. My calculation amounts to the effect of the whole sphere to the given hemisphere. Yet, since the net force applied by the hemisphere itself is zero, this gives the correct answer.

The result turned out to be,

\mathbf{F}=\frac{\pi\rho^{2}R^{4}}{12 \epsilon_0}\hat{z}

Anyone please confirm it for me.