Two spherical conductors of radii 4 cm and 5 | vedclass

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(A) Given that the two spherical conductors are at the same potential $V$.For a spherical conductor of radius $r$ and charge $q$,the potential is give

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$5$:$4$

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(A) Given that the two spherical conductors are at the same potential $V$.For a spherical conductor of radius $r$ and charge $q$,the potential is given by $V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}$.Since $V_1 = V_2$,we have $\frac{q_1}{r_1} = \frac{q_2}{r_2}$,which implies $\frac{q_1}{q_2} = \frac{r_1}{r_2}$.The surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi r^2}$.Therefore,the ratio of surface charge densities is $\frac{\sigma_1}{\sigma_2} = \frac{q_1 / (4 \pi r_1^2)}{q_2 / (4 \pi r_2^2)} = \frac{q_1}{q_2} \cdot \frac{r_2^2}{r_1^2}$.Substituting $\frac{q_1}{q_2} = \frac{r_1}{r_2}$,we get $\frac{\sigma_1}{\sigma_2} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$.Given $r_1 = 4 \ cm$ and $r_2 = 5 \ cm$,the ratio is $\frac{\sigma_1}{\sigma_2} = \frac{5}{4}$.

Two spherical conductors of radii $4 \ cm$ and $5 \ cm$ are charged to the same potential. If $\sigma_1$ and $\sigma_2$ are the respective values of the surface charge density on the two conductors,then the ratio $\sigma_1 : \sigma_2$ is

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