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(B) The electric potential $V$ of a charged conducting sphere of radius $R$ carrying charge $Q$ is given by $V = k \frac{Q}{R}$,where $k = \frac{1}{4

Vedclass · Accepted Answer

$R_1 : R_2$

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(B) The electric potential $V$ of a charged conducting sphere of radius $R$ carrying charge $Q$ is given by $V = k \frac{Q}{R}$,where $k = \frac{1}{4 \pi \epsilon_0}$.Since both spheres are charged to the same potential $V$,we have:$V_1 = V_2$$k \frac{Q_1}{R_1} = k \frac{Q_2}{R_2}$Dividing both sides by $k$,we get:$\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$Rearranging the terms to find the ratio of charges $\frac{Q_1}{Q_2}$:$\frac{Q_1}{Q_2} = \frac{R_1}{R_2}$Thus,the ratio of charges on the spheres is $R_1 : R_2$.

Two metal spheres of radii $R_1$ and $R_2$ are charged to the same potential. The ratio of charges on the spheres is

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