Charged Sphere sliced, force required to keep them as they were

[AGNuke]

A metallic sphere of radius R is cut in two parts along a plane whose minimum distance from the sphere's centre is h and the sphere is uniformly charged by a total electric charge Q. What minimum force is necessary to hold the two parts of the sphere together?

The Real trivia
The solution which we were "encouraged" to come up was using a term called "electric pressure", defined as the electrostatic force per unit area. By multiplying it with the base area of the cross-section obtained after slicing the sphere, I got the answer.

$$P_{el}=\frac{\sigma ^{2}}{2\varepsilon _{0}}; \; \sigma =\frac{Q}{4\pi R^2}$$
$$F_{el}=P_{el}\times S; \; S=\pi(R^2-h^2)$$
$$F_{el}=\frac{\frac{Q^2}{16\pi^2R^4}}{2\varepsilon_{0}}\times \pi(R^2-h^2)=\frac{Q^2(R^2-h^2)}{32\pi\varepsilon_{0}R^4}$$

Now I actually didn't get the concept here, what was that supposed to mean. Isn't there a conventional way to solve this problem?

 

[TSny]

Hi AGNuke. I'm not too clear on what specific question you are asking. Are you asking why the pressure is given by σ²/2ε_o or are you asking why the force is given by F = P*$\pi$(R²-h²)? Or are you asking something else?

The expression for the pressure can be obtained in a fairly conventional way by considering the force on a small patch of area of the surface of a charged conductor.

 

[AGNuke]

TSny, I am sorry if I was unable to convey my question properly, but I am asking that can I solve this question using coulomb's law and some other textbook stuff like properties of conductors, etc.?

I mean, if I don't know the concept of electric field (which I am still trying to justify, even if for a sphere), can I solve it? If so, then what should I do? (Coulomb's Law?)

 

[TSny]

A direct integration using Coulomb's law is a bit messy. Using the following concepts involving electric field makes it a lot easier:

(1) The relation between force and electric field: F = qE

(2) The relation between electric field at the surface of a conductor and the charge density: E = σ/ε_o

(3) The electric field produced by a large flat sheet of uniform charge density: E = σ/2ε_o

The latter 2 properties are easily derived from Gauss' law.

 

[Nickie]

I can provide a different approach to solving this problem. Instead of using the concept of electric pressure, we can use the principle of conservation of energy. The minimum force required to hold the two parts of the sphere together is equal to the change in potential energy when the sphere is sliced and brought back together.

The potential energy of a charged sphere is given by U = kQ^2/R, where k is the Coulomb constant. When the sphere is sliced, the potential energy of each half becomes U/2. When the two halves are brought back together, the potential energy becomes U again.

Therefore, the change in potential energy is U - U/2 = U/2. This change in potential energy is equal to the work done by the external force to hold the two parts together.

Thus, the minimum force required is given by F = U/2h, where h is the distance between the two parts of the sphere. Substituting the value of U, we get F = kQ^2/2Rh.

This approach provides a more intuitive understanding of the minimum force required to hold the two parts of the charged sphere together. It also avoids the use of electric pressure, which may not be a concept familiar to all scientists.