Two spherical conductors of capacities 3 mu F | vedclass

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(C) We know that the capacitance of a spherical conductor is $C = 4 \pi \epsilon_0 r$. Since $C \propto r$,the ratio of capacitances is equal to the r

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$\frac{2}{3}$

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(C) We know that the capacitance of a spherical conductor is $C = 4 \pi \epsilon_0 r$. Since $C \propto r$,the ratio of capacitances is equal to the ratio of radii: $\frac{C_1}{C_2} = \frac{r_1}{r_2} = \frac{3}{2}$.Given that both spheres are at the same potential $V$,the charges are $Q_1 = C_1 V$ and $Q_2 = C_2 V$.The surface charge density is given by $\sigma = \frac{Q}{A} = \frac{Q}{4 \pi r^2}$.Substituting $Q = CV$,we get $\sigma = \frac{CV}{4 \pi r^2}$.Since $C = 4 \pi \epsilon_0 r$,we have $\sigma = \frac{(4 \pi \epsilon_0 r) V}{4 \pi r^2} = \frac{\epsilon_0 V}{r}$.Thus,$\sigma \propto \frac{1}{r}$.Therefore,$\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1} = \frac{2 \ cm}{3 \ cm} = \frac{2}{3}$.

Capacitor Type	Capacitance Formula
$A$. Cylindrical capacitor	$i$. $4\pi \epsilon_0 R$
$B$. Spherical capacitor	$ii$. $\frac{K A \epsilon_0}{d}$
$C$. Parallel plate capacitor with dielectric	$iii$. $\frac{2\pi \epsilon_0 \ell}{\ln(r_2/r_1)}$
$D$. Isolated spherical conductor	$iv$. $\frac{4\pi \epsilon_0 r_1 r_2}{r_2 - r_1}$

Two spherical conductors of capacities $3 \mu F$ and $2 \mu F$ are charged to the same potential,having radii $3 \ cm$ and $2 \ cm$ respectively. If $\sigma_1$ and $\sigma_2$ represent the surface charge density on the respective conductors,then $\frac{\sigma_1}{\sigma_2}$ is:

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