Force of interaction between caps

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

I can extend the bigger cap with charge density $\rho_2$ and make it a complete sphere. If I do this, I must replace the smaller cap of charge density $\rho_1$ with a cap (of same dimensions) of charge density $\rho_1-\rho_2$. Now this doesn't look right to me because if I continue to find the force of interaction with the help of this new configuration, the force turns out to be attractive as $\rho_1 - \rho_2$ is negative where as the force should be repulsive.

Any help is appreciated. Thanks!

 

[TSny]

I can extend the bigger cap with charge density $\rho_2$ and make it a complete sphere. If I do this, I must replace the smaller cap of charge density $\rho_1$ with a cap (of same dimensions) of charge density $\rho_1-\rho_2$.

I'm not following you here.

Can you find the force of interaction for the case of equal densities $\rho_1 = \rho_2$?

Then try to make a simple argument for dealing with the case of unequal densities.

 

[Saitama]

I'm not following you here.

Okay, let me explain with the help of a picture.

Consider a sphere with charge density $\rho_2$ (blue). Then I place a spherical cap of charge density $\rho_1-\rho_2$ (red) on it. This is equivalent to the given setup in the problem. Can you understand my approach now? Please let me know if anything is still unclear.

 

[TSny]

OK. I'm not sure how this way of looking at it helps. The cap with density $\rho_1-\rho_2$ will be attracted to the lower section of density $\rho_2$. But don't forget that you would also need to include an additional repulsive force on a cap of density $\rho_2$ for a uniform sphere of density $\rho_2$.

[EDIT: I would start with a uniform sphere of density $\rho_1$ and find the force on the cap (of density $\rho_1$). You should then be able to tweak the answer to get the solution to the problem.]

 

[Saitama]

[EDIT: I would start with a uniform sphere of density $\rho_1$ and find the force on the cap (of density $\rho_1$). You should then be able to tweak the answer to get the solution to the problem.]

I guess you meant uniform sphere of density $\rho_2$.

I seem to be ending up with triple integrals if I try to find the force. :(

Here is my attempt:

Consider a small element inside the cap of charge density $\rho_1$ of volume:
$$dV=r\,d\theta\,dr\,r\cos\theta\, d\phi=r^2\cos\theta\,dr\,d\theta\,d\phi$$
$\phi$ is the second angle which is generally used in spherical coordinates. (I can't draw in 3D, I hope you understand what I am trying to do. )

The charge in this volume is $\rho_1\,dV$.

The electric field due to the larger sphere for $r<R$ is given by $\rho_2r/(3\epsilon_0)$. Hence, the force on the selected element is given by:
$$dF=\frac{\rho_2\,r\, \rho_1\,dV}{3\epsilon_0}$$
The above force is radially outward so I resolve it into components. By symmetry, the force is in the direction of F shown in the figure. Hence, the component of force we have to integrate is the following:
$$dF'=\frac{\rho_2 \,r\,\rho_1\,dV}{3\epsilon_0}\cos\theta \,\cos\phi$$
I am sorry TSny but I can't explain how I got the above expression. I know how to reach this expression but I just don't know how to put it in words, I can't even make a sketch to explain my thought process. Sorry again. :(

Following are the limits for integration:

For $\phi$: 0 to $\arccos(1-h/(r\cos\theta))$
For $\theta$: 0 to $\arccos(1-h/r)$
For $r$: $R-h$ to R.

Is the above correct?

Or is the above completely gibberish?

 

[TSny]

I guess you meant uniform sphere of density $\rho_2$.

I was thinking of a uniform sphere of density $\rho_1$ throughout. It was easier for me to think about this case first and then use it to get the result for case where the cap and the associated truncated sphere have different densities.

The electric field due to the larger sphere for $r<R$ is given by $\rho_2r/(3\epsilon_0)$.

OK. But note that this is the field of a uniformly charged sphere of density $\rho_2$ throughout the sphere. You need to justify why you can use this field rather than the field due to just the truncated sphere of density $\rho_2$.

Hence, the force on the selected element is given by:
$$dF=\frac{\rho_2\,r\, \rho_1\,dV}{3\epsilon_0}$$
The above force is radially outward so I resolve it into components. By symmetry, the force is in the direction of F shown in the figure. Hence, the component of force we have to integrate is the following:
$$dF'=\frac{\rho_2 \,r\,\rho_1\,dV}{3\epsilon_0}\cos\theta \,\cos\phi$$

I don't understand the factor of $\cos \phi$. I have rotated your figure. If we let the vertical axis be the z-axis, then you want the z-component of the force dF on the charge element.

Also, I don't think your expression for the volume element is correct (if you are using the standard notation for $\theta$ and $\phi$).

Consider a small element inside the cap of charge density $\rho_1$ of volume:
$$dV=r\,d\theta\,dr\,r\cos\theta\, d\phi=r^2\cos\theta\,dr\,d\theta\,d\phi$$

 

[Saitama]

Also, I don't think your expression for the volume element is correct (if you are using the standard notation for $\theta$ and $\phi$).

I will get back to other issues after this because I feel I must clear the confusion in the coordinate system I am using.

I am not using the standard notation for $\theta$. My $\theta$ is different. I know that in spherical coordinates, my expression for dV should have sin instead of cos but I am not comfortable with using that. If $\theta$ is the angle in spherical coordinates, my $\theta$ is $\pi/2-\theta$ which is the reason for factor of $\cos\theta$. Can you follow my working now?

I am sorry for creating so much confusion. I rarely use these different type of coordinate systems and triple integral so I tend to mess up.

 

[TSny]

I still don't understand why you put a factor of $\cos \phi$ in your expression for $dF'$. The force is radial and the radial direction makes an angle $\theta$ to the z-axis. So, shouldn't the z-component of $dF'$ be $dF' \cos \theta$?

[EDIT: I guess you are taking the x-axis rather than the z-axis as the axis of symmetry for the cap. This seems very confusing to me.]

 

[Saitama]

Its getting very difficult for me to explain my thought process so I am changing the method.

This time, I am thinking of dividing the cap of charge density $\rho_1$ into cylinders (or rings). Please have a look at the attachment.

Consider a cylinder of thickness $dr$ and height $dx$. The charge in this cylinder is $\rho_1dV$ where $dV=2\pi r\,dr\,dx$. From geometry,
$$r=x\tan\theta \Rightarrow dr=\tan \theta\,dx+x\sec^2\theta\,d\theta$$
Each part of the cylinder is at a distance $x\sec\theta$ from the centre.

The force on the selected cylinder in the direction of F (right) shown in the sketch is:
$$dF'=\frac{\rho_2 x\sec\theta\rho_1\,dV}{3\epsilon_0}\cos\theta=\frac{\rho_2 x\rho_1\,dV}{3\epsilon_0}$$
Does this approach sound better? I am sorry if this is getting more confusing.

Let me clear something, I am finding the force due to a complete sphere of charge density $\rho_2$ on the cap of charge density $\rho_1$.

EDIT: I feel I don't need to express $r$ in terms of $\theta$ and $x$. Can I use the following integration limits:

For $r$: from r=0 to $\sqrt{R^2-x^2}$
For $x$: from x=R-h to R

 

[TSny]

The force on the selected cylinder in the direction of F (right) shown in the sketch is:
$$dF'=\frac{\rho_2 x\sec\theta\rho_1\,dV}{3\epsilon_0}\cos\theta=\frac{\rho_2 x\rho_1\,dV}{3\epsilon_0}$$
Does this approach sound better?

Yes, that looks very good.

the following integration limits:

For $r$: from r=0 to $\sqrt{R^2-x^2}$
For $x$: from x=R-h to R

Is the lower limit on x correct?

 

[Saitama]

Yes, that looks very good except for the lower limit on x.

Sorry about that, it should be h.

I continue with my expression of dF'.

$$dF'=\frac{\rho_1\rho_2}{3\epsilon_0}x\,dV=\frac{\rho_1 \rho_2}{3\epsilon_0}2\pi x r\,dr\,dx$$
$$\Rightarrow F'=\frac{2\pi \rho_1 \rho_2}{3\epsilon_0}\int_h^R \int_0^{\sqrt{R^2-x^2}} xr\,dr\,dx$$
$$\Rightarrow F'=\frac{2\pi \rho_1 \rho_2}{3\epsilon_0}\int_h^R \frac{x(R^2-x^2)}{2}\,dx$$
$$\Rightarrow F'=\frac{\pi \rho_1 \rho_2}{3\epsilon_0}\frac{R^4}{4}$$

What should I do next?

 

[TSny]

$$\Rightarrow F'=\frac{\pi \rho_1 \rho_2}{3\epsilon_0}\frac{R^4}{4}$$

I don't get the same numerical coefficient as you do. Otherwise, OK.

I agree with the following
$$\Rightarrow F'=\frac{2\pi \rho_1 \rho_2}{3\epsilon_0}\int_h^R \frac{x(R^2-x^2)}{2}\,dx$$

 

[Saitama]

I agree with the following
$$\Rightarrow F'=\frac{2\pi \rho_1 \rho_2}{3\epsilon_0}\int_h^R \frac{x(R^2-x^2)}{2}\,dx$$

Sorry again.

This time I get:
$$F'=\frac{\pi \rho_1 \rho_2}{3\epsilon_0}\frac{(R^2-h^2)^2}{4}$$

Is this correct?

What should my next step?

 

[TSny]

Sorry again.

This time I get:
$$F'=\frac{\pi \rho_1 \rho_2}{3\epsilon_0}\frac{(R^2-h^2)^2}{4}$$

Is this correct?

Looks correct.

What should my next step?

Interpret what you have calculated. Is it the answer to the problem? Why or why not?

 

[Saitama]

Interpret what you have calculated. Is it the answer to the problem? Why or why not?

No, it is not the answer to the problem because the problem asks the force between two caps, not between a sphere and a cap.

I am thinking the following:

Since I added an extra cap of charge density $\rho_2$ to make the larger cap a complete sphere, I must place a similar cap of charge density $-\rho_2$ so that the original configuration is retained. So for the final answer , I should also add the attractive force between the sphere and the cap of charge density $-\rho_2$. The following is the expression I get:
$$F=\frac{\pi \rho_2}{3\epsilon_0}\frac{(R^2-h^2)^2}{4}\left( \rho_1-\rho_2^2\right)$$
Is this correct?

 

[TSny]

$$F=\frac{\pi \rho_2}{3\epsilon_0}\frac{(R^2-h^2)^2}{4}\left( \rho_1-\rho_2^2\right)$$
Is this correct?

No, I don't believe that's right. (The square on $\rho_2$ inside the parentheses is a typographical error, but it still won't be right if you correct this error.)

The extra term you have added represents the force that a full sphere of density $\rho_2$ exerts on a cap of density -$\rho_2$. But that's not what you want.

Go back to $F'=\frac{\pi \rho_1 \rho_2}{3\epsilon_0}\frac{(R^2-h^2)^2}{4}$. As you noted, this represents the force that a full sphere of density $\rho_2$ exerts on the cap of density $\rho_1$. You want the force, $F$, that the lower truncated sphere of density $\rho_2$ exerts on the cap of density $\rho_1$.

How would you interpret the difference $F'-F$ in words?

 

[Saitama]

How would you interpret the difference $F'-F$ in words?

I am not sure but is it the force which acts upon $\rho_1$ due to the extra $\rho_2$ I introduced?

 

[TSny]

Yes. $F'-F$ is the force on the cap of density $\rho_1$ due to charge of density $\rho_2$ occupying the same cap region.

Think about how you might deduce what that force is without doing any complicated integration.

 

[Saitama]

Think about how you might deduce what that force is without doing any complicated integration.

I can't think of anything and I don't see a way to set up the integrals here. :(

 

[TSny]

First suppose you had just the cap isolated by itself and it had a uniform charge density $\rho_0$. What would be the net electric force on the cap (due to all the charge elements within the cap exerting electric force on each other)?

 

[Saitama]

First suppose you had just the cap isolated by itself and it had a uniform charge density $\rho_0$. What would be the net electric force on the cap (due to all the charge elements within the cap exerting electric force on each other)?

I have no idea about how you would calculate it and setting up the integrals would be very messy.

 

[TSny]

Consider any system of point charges. Each particle exerts an electric force on each of the other particles. If you add up all of the forces on all of the particles to get the net force on the entire system, what do you get?

 

[Saitama]

Consider any system of point charges. Each particle exerts an electric force on each of the other particles. If you add up all of the forces on all of the particles to get the net force on the entire system, what do you get?

Umm...isn't that the same question you asked me in #21?

 

[haruspex]

Umm...isn't that the same question you asked me in #21?

Yes, just generalized, in the hope of making the answer more obvious.
Think of it this way... the particles in a spaceship exert various forces on each other. The sum of all these forces is F. F=ma. What is the resultant acceleration?

 

[Saitama]

Yes, just generalized, in the hope of making the answer more obvious.
Think of it this way... the particles in a spaceship exert various forces on each other. The sum of all these forces is F. F=ma. What is the resultant acceleration?

a=F/m?

I know the above sounds stupid. I still don't see how to find this force.

 

[TSny]

The key is Newton's third law which holds for electrostatic force. If you had a system of just two positive point charges, they would want to fly away from each other. But the forces have equal magnitude and opposite direction. So, the net electric force on the system is zero.

If the charges are fixed to the ends of a rod so they can't fly away, then the electric force of each charge on the other still obeys the 3rd law.The net electric force is still zero. If the system starts at rest, it remains at rest. The system can't "pull itself up by its own bootstraps".

Generalize this to any system of point charges. In particular, what about a charged cap of the sphere. How much electric force does it exert on itself?

 

[haruspex]

a=F/m?

I know the above sounds stupid. I still don't see how to find this force.

I meant this as a thought experiment: what acceleration would you expect a spaceship to achieve arising entirely from the internal forces between its components?

 

[haruspex]

Is this still of interest? It's quite a pretty problem, so I'd hate to see it atrophy.
Call the regions occupied by the caps A, B (A being the smaller), and call the complete sphere S.
If a body with charge ρ₀ occupies A then call it A₀ etc., so the two bodies are A₀, B₁.
To make the integral easier, it would be convenient to consider the force between A₀ and S₁. To that end, we need to calculate that between A₀ and A₁.
Consider two small regions X, Y in A. X₁ exerts a force on Y₀, while Y₁ exerts an equal and opposite force on X₀. Thus, there is no net force between A₀ and A₁.
It follows that the force between A₀ and B₁ is the same as that between A₀ and S₁.
Using cylindrical coordinates with origin at the centre of S, and the z axis orthogonal to the A/B boundary, consider a point in A at (r, θ, z). What is the field at that point due to S₁? What is the component of that parallel to the z axis?

 

[TSny]

Is this still of interest? It's quite a pretty problem, so I'd hate to see it atrophy.
Call the regions occupied by the caps A, B (A being the smaller), and call the complete sphere S.
If a body with charge ρ₀ occupies A then call it A₀ etc., so the two bodies are A₀, B₁.
To make the integral easier, it would be convenient to consider the force between A₀ and S₁. To that end, we need to calculate that between A₀ and A₁.

That's a nice recap (sorry for the pun) of the approach we were trying to take. We had gotten to the problem of finding the force of A₁ on A₀. But this is where Pranav seems to have either lost interest or been too busy with other things to continue.

Consider two small regions X, Y in A. X₁ exerts a force on Y₀, while Y₁ exerts an equal and opposite force on X₀. Thus, there is no net force between A₀ and A₁.

That's a nice and very succinct argument.

It follows that the force between A₀ and B₁ is the same as that between A₀ and S₁.

Yes, that's what I was hoping to get Pranav to see.

Using cylindrical coordinates with origin at the centre of S, and the z axis orthogonal to the A/B boundary, consider a point in A at (r, θ, z). What is the field at that point due to S₁? What is the component of that parallel to the z axis?

I think Pranav did this calculation and his result is in post #13, although he chose the x-axis to be the symmetry axis rather than the z-axis.

Thanks for your contribution to keep this thread moving! I agree it's a nice problem.

 

[haruspex]

I think Pranav did this calculation and his result is in post #13, although he chose the x-axis to be the symmetry axis rather than the z-axis.

OK, but I disagree with the answer in that post. I don't see any expression involving the volume of S. I get a pi-squared in the answer.