Two conducting spheres of radii r_1 and r_2 h | vedclass

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Question

Electric field on the surface of conducting sphere $=\frac{k Q}{R^{2}}$

according to given condition

$\frac{k q_{1}}{r_{1}^{2}}=\frac{k q_{2}}{r

Vedclass · Accepted Answer

$(r_1/r_2)$

Vedclass · Answer

Electric field on the surface of conducting sphere $=\frac{k Q}{R^{2}}$

according to given condition

$\frac{k q_{1}}{r_{1}^{2}}=\frac{k q_{2}}{r_{2}^{2}}$

$\Longrightarrow \frac{q_{1}}{q_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}-(1)$

ratio of potentials at centre

$=\frac{\frac{k q_{1}}{r_{1}}}{\frac{k q_{2}}{r_{2}}}=\frac{q_{1}}{q_{2}} \cdot \frac{r_{2}}{r_{1}}$

substituting $\frac{q_{1}}{q_{2}}$ from $( 1)$

we get

ratio $=\frac{r_{1}}{r_{2}}$ Ans.

Two conducting spheres of radii $r_1$ and $r_2$ have same electric fields near their surfaces. The ratio of their electric potentials is

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