Two spheres A and B of radius a and b respect | vedclass

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(B) Given that the electric potential of both spheres is the same,i.e.,$V_A = V_B$.Using the formula for the potential of a charged sphere $V = \frac{

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$b/a$

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(B) Given that the electric potential of both spheres is the same,i.e.,$V_A = V_B$.Using the formula for the potential of a charged sphere $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r}$,we have:$\frac{1}{4\pi\epsilon_0} \cdot \frac{Q_A}{a} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_B}{b} \implies \frac{Q_A}{Q_B} = \frac{a}{b}$......$(i)$Surface charge density $\sigma$ is defined as $\sigma = \frac{Q}{4\pi r^2}$.Therefore,the ratio of surface charge densities is:$\frac{\sigma_A}{\sigma_B} = \frac{Q_A}{4\pi a^2} \cdot \frac{4\pi b^2}{Q_B} = \frac{Q_A}{Q_B} \cdot \frac{b^2}{a^2}$Substituting the ratio from equation $(i)$:$\frac{\sigma_A}{\sigma_B} = \frac{a}{b} \cdot \frac{b^2}{a^2} = \frac{b}{a}$.

Two spheres $A$ and $B$ of radius $a$ and $b$ respectively are at the same electric potential. The ratio of the surface charge densities of $A$ and $B$ is

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