Why is the pressure on a charged sphere only exerted on the contact area?

[phantomvommand]

Homework Statement

A metal sphere of radius R is cut in 2 along a plane whose minimum distance from the sphere's centre is h. The sphere has a uniform charge of Q. What force is necessary to hold the 2 parts of the sphere together?

Relevant Equations

F = 1/2 E Q

The force per unit area (Pressure) on a part of the sphere is given by F = (E outside + E inside)/2 * Q = 0.5 (kQ/R^2) * (Q/ 4piR^2) = (Q^2/ 32pi^2 e0 R^4).

I understand the above line.

The solution then says this pressure is exerted on the contact area between the 2 spheres, as given by pi(R^2 - h^2).
Total force needed is Pressure x Area.

Why that the pressure only exerted on the contact area? Isn't that the pressure exerted on the surface of the sphere, and so what is the link between the pressure on the surface of the sphere being equal to the pressure on the contact area?

 

[haruspex]

It says metal sphere, so is this a uniform surface charge?

 

[phantomvommand]

It says metal sphere, so is this a uniform surface charge?

yes, that is why we are using surface charge density as given by Q/4piR^2 in the calculation of surface pressure. Why is the surface pressure equivalent to the pressure exerted on the contact surface between the 2 parts of the sliced sphere?

 

[haruspex]

yes, that is why we are using surface charge density as given by Q/4piR^2 in the calculation of surface pressure. Why is the surface pressure equivalent to the pressure exerted on the contact surface between the 2 parts of the sliced sphere?

The force exerted by the pressure on a small area element is normal to the element. For the net force over the whole cap you need to integrate this as a vector. The components normal to the axis of the cap will cancel.

 

[phantomvommand]

The force exerted by the pressure on a small area element is normal to the element. For the net force over the whole cap you need to integrate this as a vector. The components normal to the axis of the cap will cancel.

that could be one way. But the solution states that the net force would simply be

Pressure x contact area,

why is that so?

 

[haruspex]

that could be one way. But the solution states that the net force would simply be

Pressure x contact area,

why is that so?

For the reason I gave. $F=\int P.\vec{dA}=P\int\vec{dA}$.
If the plane of the cut is S (represented by a vector $\vec S$ normal to the plane S) then each $\vec{dA}$ can be resolved into an area element vector parallel to vector $\vec S$ and a vector normal to $\vec S$.
In the integral, the components normal to $\vec S$ cancel, leaving only the projection of the cap onto S.

 

[phantomvommand]

For the reason I gave. $F=\int P.\vec{dA}=P\int\vec{dA}$.
If the plane of the cut is S (represented by a vector $\vec S$ normal to the plane S) then each $\vec{dA}$ can be resolved into an area element vector parallel to vector $\vec S$ and a vector normal to $\vec S$.
In the integral, the components normal to $\vec S$ cancel, leaving only the projection of the cap onto S.

thank you!