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Question

(A) The electrostatic pressure on the surface of a charged conductor is given by $P = \frac{\sigma^2}{2\varepsilon_0}$.This pressure acts radially out

Vedclass · Accepted Answer

$\frac{1}{\varepsilon_0} \sigma^2 R^2$

Vedclass · Answer

(A) The electrostatic pressure on the surface of a charged conductor is given by $P = \frac{\sigma^2}{2\varepsilon_0}$.This pressure acts radially outward on every element of the surface.To keep the two hemispherical shells together, the external force $F$ must balance the total outward force exerted by the electrostatic pressure on the cross-sectional area of the hemisphere.The cross-sectional area of the hemisphere is $A = \pi R^2$.The total outward force $F$ is the product of the electrostatic pressure and the projected area perpendicular to the force direction:$F = P 	imes A = \left( \frac{\sigma^2}{2\varepsilon_0} ight) 	imes \pi R^2$.Thus, $F = \frac{\pi \sigma^2 R^2}{2\varepsilon_0}$.Since $\pi$, $2$, and $\varepsilon_0$ are constants, the force $F$ is proportional to $\sigma^2 R^2$.

$A$ uniformly charged thin spherical shell of radius $R$ carries a uniform surface charge density of $\sigma$ per unit area. It is made of two hemispherical shells, held together by pressing them with force $F$ (see figure). $F$ is proportional to

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