- #1
AGNuke
Gold Member
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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?
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?
