If the potential on the surface of two conducting spheres of radii $r_{1}$ and $r_{2}$ are same then the ratio of their respective surface charge densities will be

r_{1} : r_{2}

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r_{2} : r_{1}

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1 : 1

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r_{1}^{2} : r_{2}^{2}

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Solution

The correct option is B $r_{2} : r_{1}$
$C_{1} = 4 π ε_{o} r_{1}$
$C_{2} = 4 π ε_{o} r_{2}$
After connections
find potential is v
$Q_{1} - C_{1} V = 4 π ε_{o} r_{1} v$
$Q_{2} - C_{2} V = 4 π ε_{o} r_{2} v$
$\frac{Q_{1}}{Q_{2}} = \frac{4 π ε_{o} r_{1} v}{4 π ε_{o} r_{2} v} = \frac{r_{1}}{r_{2}}$ …………..(i)
$\frac{E_{1}}{E_{2}} = \frac{\frac{k Q_{1}}{r_{1}^{2}}}{\frac{k Q_{2}}{r_{2}^{2}}} = (\frac{Q_{1}}{Q_{2}}) \frac{(r_{2})^{2}}{(r_{1})^{2}}$ …………(ii)
From (i) & (ii)
$\frac{E_{1}}{E_{2}} = \frac{r_{2}}{r_{1}}$ .

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Similar questions

Two conducting spheres of radii $r_{1}$ and r₂ are at the same potential. Their charges are in the ratio of

r_{1}

r_{2}

r_{1}

r_{2}

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

R_{1}

R_{2}

R_{1} > R_{2}

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