Two spherical conductors A and B of radii 1 m | vedclass

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Question

After connection, $V_{1}=V_{2}$

$\Rightarrow K \frac{Q_{1}}{r_1}=K \frac{Q_{2}}{r_{2}}$

$\Rightarrow \frac{Q_{1}}{r_{1}}=\frac{Q_{2}}{r_{2}}$

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

After connection, $V_{1}=V_{2}$

$\Rightarrow K \frac{Q_{1}}{r_1}=K \frac{Q_{2}}{r_{2}}$

$\Rightarrow \frac{Q_{1}}{r_{1}}=\frac{Q_{2}}{r_{2}}$

The ratio of electric fields

$\frac{E_{1}}{E_{2}}=\frac{K \frac{Q_{1}}{r_{1}^{2}}}{K \frac{Q_{2}}{r_{2}^{2}}}=\frac{Q_{1}}{r_1^{2}} 	imes \frac{r_{2}^{2}}{Q_{2}}$

$\Rightarrow \frac{E_{1}}{E_{2}}=\frac{r_{1} 	imes r_{2}^{2}}{r_1^{2} 	imes r_{2}}$

$ \Rightarrow \frac{E_{1}}{E_{2}}=\frac{r_{2}}{r_{1}}=\frac{2}{1}$

since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

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