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(D) The electrostatic pressure on the surface of a charged conductor is given by $P = \frac{\sigma^2}{2\epsilon_0}$,where $\sigma$ is the surface char

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$\frac{Q^2(R^2 - h^2)}{32\pi \epsilon_0 R^4}$

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(D) The electrostatic pressure on the surface of a charged conductor is given by $P = \frac{\sigma^2}{2\epsilon_0}$,where $\sigma$ is the surface charge density.Given the total charge $Q$ on a sphere of radius $R$,the surface charge density is $\sigma = \frac{Q}{4\pi R^2}$.Thus,the pressure is $P = \frac{1}{2\epsilon_0} \left( \frac{Q}{4\pi R^2} ight)^2 = \frac{Q^2}{32\pi^2 \epsilon_0 R^4}$.The force required to hold the two parts together is the integral of the pressure component normal to the cross-sectional plane. This is equivalent to the pressure multiplied by the area of the cross-section.The radius of the circular cross-section formed by the plane is $r = \sqrt{R^2 - h^2}$.The area of this cross-section is $A = \pi r^2 = \pi(R^2 - h^2)$.The total force $F$ is $F = P 	imes A = \left( \frac{Q^2}{32\pi^2 \epsilon_0 R^4} ight) 	imes \pi(R^2 - h^2) = \frac{Q^2(R^2 - h^2)}{32\pi \epsilon_0 R^4}$.

Consider a metal sphere of radius $R$ that is cut into two parts along a plane whose minimum distance from the sphere's centre is $h$. The sphere is uniformly charged with a total electric charge $Q$. The minimum force necessary to hold the two parts of the sphere together is:

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