Given below are two statements: One is labell | vedclass

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(A) The potential $V$ of a conducting sphere of radius $R$ carrying charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.This formula app

Vedclass · Accepted Answer

$A$ is false but $R$ is true

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(A) The potential $V$ of a conducting sphere of radius $R$ carrying charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.This formula applies to both solid and hollow conducting spheres because the charge resides entirely on the outer surface.Given that both spheres have the same radius $R$ and are charged to the same potential $V$,we have $V = \frac{Q_1}{4\pi\epsilon_0 R}$ and $V = \frac{Q_2}{4\pi\epsilon_0 R}$.Equating these,we get $Q_1 = Q_2$.Therefore,the solid sphere and the hollow sphere will have the same charge,making Assertion $A$ false.Reason $R$ states that the capacitance of a metallic sphere depends on its radius $(C = 4\pi\epsilon_0 R)$,which is a true statement.Thus,$A$ is false but $R$ is true.

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