(D) The potential $V$ of a conducting sphere of radius $r$ with surface charge density $\sigma$ is given by $V = \frac{\sigma r}{\varepsilon_0}$.When

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(D) The potential $V$ of a conducting sphere of radius $r$ with surface charge density $\sigma$ is given by $V = \frac{\sigma r}{\varepsilon_0}$.When two conducting spheres are brought into contact,charge flows between them until their potentials become equal,i.e.,$V_1 = V_2$.Substituting the expression for potential,we get $\frac{\sigma_1 r_1}{\varepsilon_0} = \frac{\sigma_2 r_2}{\varepsilon_0}$.This simplifies to $\sigma_1 r_1 = \sigma_2 r_2$.Given $r_1 = R$ and $r_2 = 3R$,we have $\sigma_1 R = \sigma_2 (3R)$.Therefore,the ratio $\frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3$.

Class 12 Physics Electric Potential and Capacitance Conductor, Electrostatic Shielding, Induced Charge and Charge Redistribution on conductor

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Two metal spheres of radius $R$ and $3R$ have the same surface charge density $\sigma$. If they are brought in contact and then separated,the surface charge density on the smaller and bigger sphere becomes $\sigma_1$ and $\sigma_2$,respectively. The ratio $\frac{\sigma_1}{\sigma_2}$ is:

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A
$\frac{1}{9}$
B
$9$
C
$\frac{1}{3}$
D
$3$

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Electric Potential and Capacitance

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Conductor, Electrostatic Shielding, Induced Charge and Charge Redistribution on conductor

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Two metal spheres of radius $R$ and $3R$ have the same surface charge density $\sigma$. If they are brought in contact and then separated,the surface charge density on the smaller and bigger sphere becomes $\sigma_1$ and $\sigma_2$,respectively. The ratio $\frac{\sigma_1}{\sigma_2}$ is:

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Two metal spheres $A$ and $B$ of radii $a$ and $b$ $(a < b)$ respectively are at a large distance apart. Each sphere carries a charge of $100 \mu C$. The spheres are connected by a conducting wire,then

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