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(B) The electric pressure $P$ on the surface of a charged conductor is given by $P = \frac{\sigma^2}{2\varepsilon_0}$,where $\sigma$ is the surface ch

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$F = \frac{Q^2}{32\pi \varepsilon_0 R^2}$

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(B) The electric pressure $P$ on the surface of a charged conductor is given by $P = \frac{\sigma^2}{2\varepsilon_0}$,where $\sigma$ is the surface charge density.Given the total charge $Q$ is uniformly distributed over the sphere of radius $R$,the surface charge density is $\sigma = \frac{Q}{4\pi R^2}$.The force $F$ required to hold the two hemispheres together is equal to the net electrostatic repulsive force acting on one hemisphere due to the other. This force is equivalent to the electric pressure acting on the projected area of the hemisphere.The projected area of a hemisphere of radius $R$ is $A = \pi R^2$.Thus,the force $F = P 	imes A = \left(\frac{\sigma^2}{2\varepsilon_0}ight) 	imes (\pi R^2)$.Substituting $\sigma = \frac{Q}{4\pi R^2}$:$F = \frac{1}{2\varepsilon_0} \left(\frac{Q}{4\pi R^2}ight)^2 	imes \pi R^2$$F = \frac{1}{2\varepsilon_0} 	imes \frac{Q^2}{16\pi^2 R^4} 	imes \pi R^2$$F = \frac{Q^2}{32\pi \varepsilon_0 R^2}$.

Two identical non-conducting thin hemispherical shells,each of radius $R$,are brought into contact to form a complete sphere. If a total charge $Q$ is uniformly distributed on them,what is the minimum force $F$ required to hold them together?

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