# Force of interaction between caps

1. Mar 30, 2014

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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!

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• ###### combined caps2.png
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2. Mar 30, 2014

### TSny

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.

3. Mar 30, 2014

### Saitama

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.

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• ###### caps.png
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4. Mar 30, 2014

### 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.]

Last edited: Mar 31, 2014
5. Mar 31, 2014

### Saitama

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? :uhh:

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• ###### capsss.png
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6. Mar 31, 2014

### TSny

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.

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 $\rho_2$.

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 $\phi$).

#### Attached Files:

• ###### Cap force 3.png
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Last edited: Mar 31, 2014
7. Mar 31, 2014

### Saitama

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.

8. Mar 31, 2014

### 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.]

9. Apr 1, 2014

### 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

#### Attached Files:

• ###### caps3.png
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Last edited: Apr 1, 2014
10. Apr 1, 2014

### TSny

Yes, that looks very good.

Is the lower limit on x correct?

Last edited: Apr 1, 2014
11. Apr 1, 2014

### Saitama

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?

12. Apr 1, 2014

### TSny

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

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

13. Apr 1, 2014

### Saitama

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?

14. Apr 1, 2014

### TSny

Looks correct.

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

15. Apr 1, 2014

### Saitama

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?

16. Apr 1, 2014

### TSny

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?

17. Apr 1, 2014

### Saitama

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

18. Apr 1, 2014

### 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.

19. Apr 1, 2014

### Saitama

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

20. Apr 1, 2014

### 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)?

Last edited: Apr 1, 2014