Please help with this suggested solution involving electrostatic press

[tellmesomething]

Homework Statement

Find the force of interaction between two equal halves of a non conducting uniform solid sphere of charge density $\rho$ and radius R.

Relevant Equations

None

My teacher gave us this idea:
He told us that the contact force due to one halve on the other halve and the electrostatic force will balance each other.
Therefore they'll be same in magnitude so use the former to get the answer

However i do not understand how they will be equal. Intuitively it makes sense since there are the only two forces acting on that halve therefore they should be equal and opposite to each other for them to cancel out and for the net force to be 0. But isnt this the direction of the forces : E=electrostatic N=normal force.

 

[tellmesomething]

Once im clear with this I have a doubt in the follow up solution too


Next he uses this derived result for electrostatic pressure inside a non conducting solid sphere $$P(r)= \frac{\rho²r²} {6\epsilon_{0}}$$to find the said contact force
$$F=\int P(r) 2πrdr$$
Limits are from 0 to R (sorry I forgot how to put limits in latex equations)

This is the base hes talking about

I really dont think I understand why we did this...shouldnt it be integrated over the inner part of the half sphere/hemisphere

This doesnt make a lot of sense to me....

 

[TSny]

But isnt this the direction of the forces : E=electrostatic N=normal force.

No. If you try to pull the two ends of a rod away from each other, tension builds up in the rod. At a cross-section of the rod, what direction does the material lying just to the left of the cross-section exert on the material just to the right of the cross-section?

 

[haruspex]

Find the force of interaction between two equal halves of a non conducting uniform solid sphere of charge density $\rho$ and radius R.

But it is not actually cut in half. Would it have been clearer if it had said "stuck together"?

Next he uses this derived result for electrostatic pressure inside a non conducting solid sphere $$P(r)= \frac{\rho²r²} {6\epsilon_{0}}$$to find the said contact force
$$F=\int P(r) 2πrdr$$
Limits are from 0 to R (sorry I forgot how to put limits in latex equations)

_{lower limit}^{upper limit}

shouldnt it be integrated over the inner part of the half sphere/hemisphere

Suppose you have a fluid within which the pressure P is a function of position. If you were to put a small lamina $\vec{dA }$ within the fluid the pressure on one side would be roughly constant over the lamina, leading to a force $P\vec{dA }$. Over a larger area, you can integrate that, $\int P\vec{dA }$.
(Integrating pressure over a volume gives internal energy.)

In the present case, it is a negative pressure.

 

[TSny]

One interpretation of the pressure $P(r)$ given in the problem is that it represents the effective force per unit area corresponding to the tension (contact) force between the two hemispheres.

Then, it makes more sense that your instructor says to consider the total contact force between the hemispheres.

 

[tellmesomething]

No. If you try to pull the two ends of a rod away from each other, tension builds up in the rod. At a cross-section of the rod, what direction does the material lying just to the left of the cross-section exert on the material just to the right of the cross-section?

Right, it will try to pull it so it would be away from the right cross section.. Makes sense. thankyou

 

[tellmesomething]

But it is not actually cut in half. Would it have been clearer if it had said "stuck together"?

_{lower limit}^{upper limit}

Suppose you have a fluid within which the pressure P is a function of position. If you were to put a small lamina $\vec{dA }$ within the fluid the pressure on one side would be roughly constant over the lamina, leading to a force $P\vec{dA }$. Over a larger area, you can integrate that, $\int P\vec{dA }$.
(Integrating pressure over a volume gives internal energy.)

In the present case, it is a negative pressure.

Hmm so the little laminas in the entire volume when aligned with the pressure direction, can be projected onto the base of the hemisphere? And hence we take elements along the base area?

 

[haruspex]

so the little laminas in the entire volume

No, just elements of the plane between the two hemispheres. What is happening elsewhere in the volume is irrelevant.

 

[tellmesomething]

No, just elements of the plane between the two hemispheres. What is happening elsewhere in the volume is irrelevant.

But the formular assumes radial distance from the centre? so aren't dA areas in the volume also facing electrostatic pressure?

 

[haruspex]

But the formular assumes radial distance from the centre? so aren't dA areas in the volume also facing electrostatic pressure?

Yes, there is a distribution of pressure throughout the volume, but if we notionally divide the volume into regions then the force one region exerts on a neighbour depends only on the pressure integral over the shared surface.

 

[tellmesomething]

Yes, there is a distribution of pressure throughout the volume, but if we notionally divide the volume into regions then the force one region exerts on a neighbour depends only on the pressure integral over the shared surface.

I'm not sure i understand fully. Can you suggest a way to visualise this? Should I think of the "regions" as multiple hemispherical shells?

 

[haruspex]

I'm not sure i understand fully. Can you suggest a way to visualise this? Should I think of the "regions" as multiple hemispherical shells?

I withdraw post #10. I need to think about it some more.

 

[tellmesomething]

I withdraw post #10. I need to think about it some more.

OK

 

[TSny]

To understand the formula $P(r) = \dfrac{\rho^2 r^2}{6 \varepsilon_0}$that is given in the problem, consider the total electric force on a hemispherical shell of radius $r$ and thickness $dr$ within the right hemisphere of the uniformly charged sphere of radius $R$.

We integrate $\vec E dq$ over all charge elements $dq$ within the hemispherical shell. For $\vec E$ we can use the net electric field at $dq$ due to all of the charge within the complete sphere of radius $R$; namely, $\vec E = \dfrac{\rho r}{3 \varepsilon_0} \hat r$.

When we use this total electric field in $\vec E dq$ for the force on $dq$, the force will include contributions from other elements of charge $dq’$ within the shell. However, when integrating over the complete shell, these “internal” forces will cancel out (action-reaction pairs). So, the result will represent the net force on the shell due to all charges external to the shell.

Introduce spherical coordinates $r, \theta, \phi$ as shown in the first figure above and integrate to find the total electric force on the hemispherical shell: $$F_{shell} = \frac{\pi \rho^2 r^3 dr}{3 \varepsilon_0} \hat x$$
You can now integrate over $r$ to get the net electric force on the right hemisphere.

However, we want to understand the formula $P(r) = \dfrac{\rho^2 r^2}{6 \varepsilon_0}$ given in the problem. So, stay with the hemispherical shell. The shell is in static equilibrium, so the contact force between the shell and the left half of the sphere must balance the electric force on the shell. So, the contact force on the shell will be to the left (negative-x direction). The contact region is the green ring of area $2 \pi r dr$.

The contact force per unit area is $P(r) = F_{\rm contact}/(\textrm{contact area})= F_{\textrm{shell}}/(\textrm{ring area})$ which reduces to $P(r) = \dfrac{\rho^2 r^2}{6 \varepsilon_0}$.

Thus, if you are given this formula, you can get the total electric force on the right hemisphere by integrating $P(r) dA_\rm {ring}$ over all the green rings that make up the total contact area between the left and right hemispheres.

 

[tellmesomething]

To understand the formula $P(r) = \dfrac{\rho^2 r^2}{6 \varepsilon_0}$that is given in the problem, consider the total electric force on a hemispherical shell of radius $r$ and thickness $dr$ within the right hemisphere of the uniformly charged sphere of radius $R$.

View attachment 348552

We integrate $\vec E dq$ over all charge elements $dq$ within the hemispherical shell. For $\vec E$ we can use the net electric field at $dq$ due to all of the charge within the complete sphere of radius $R$; namely, $\vec E = \dfrac{\rho r}{3 \varepsilon_0} \hat r$.

When we use this total electric field in $\vec E dq$ for the force on $dq$, the force will include contributions from other elements of charge $dq’$ within the shell. However, when integrating over the complete shell, these “internal” forces will cancel out (action-reaction pairs). So, the result will represent the net force on the shell due to all charges external to the shell.

View attachment 348553

Introduce spherical coordinates $r, \theta, \phi$ as shown in the first figure above and integrate to find the total electric force on the hemispherical shell: $$F_{shell} = \frac{\pi \rho^2 r^3 dr}{3 \varepsilon_0} \hat x$$
You can now integrate over $r$ to get the net electric force on the right hemisphere.

However, we want to understand the formula $P(r) = \dfrac{\rho^2 r^2}{6 \varepsilon_0}$ given in the problem. So, stay with the hemispherical shell. The shell is in static equilibrium, so the contact force between the shell and the left half of the sphere must balance the electric force on the shell. So, the contact force on the shell will be to the left (negative-x direction). The contact region is the green ring of area $2 \pi r dr$.
View attachment 348554


The contact force per unit area is $P(r) = F_{\rm contact}/(\textrm{contact area})= F_{\textrm{shell}}/(\textrm{ring area})$ which reduces to $P(r) = \dfrac{\rho^2 r^2}{6 \varepsilon_0}$.

Thus, if you are given this formula, you can get the total electric force on the right hemisphere by integrating $P(r) dA_\rm {ring}$ over all the green rings that make up the total contact area between the left and right hemispheres.

Sorry for the late reply! This is very nice and detailed analysis I will think on this. Thankyou so much.