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Equipotential Surfaces

The radius of...

Question

The radius of two metallic sphere $A$ and $B$ are $r_{1}$ and $r_{2}$ respectively $(r_{1} > r_{2})$ . They are connected by a thin wire and the system is given a certain charge. The charge will be greater

Equal to both

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Zero on both

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On the surface of sphere

A

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On the surface of sphere

B

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Solution

The correct option is B On the surface of sphere $A$
When the spheres are connected by a wire, then potential on both becomes equal
$∴ \frac{Q_{1}}{r_{2}} = \frac{Q_{1}}{r_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}$
when $r_{1} > r_{2}$ then $Q_{1} > Q_{2}$

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The radius of two metallic sphere A and B are r1 and r2 respectively (r1>r2). They are connected by a thin wire and the system is given a certain charge. The charge will be greater

The correct option is B On the surface of sphere AWhen the spheres are connected by a wire, then potential on both becomes equal∴Q1r2=Q1r2⇒Q1Q2=r1r2when r1>r2 then Q1>Q2

The radius of two metallic sphere $A$ and $B$ are $r_{1}$ and $r_{2}$ respectively $(r_{1} > r_{2})$ . They are connected by a thin wire and the system is given a certain charge. The charge will be greater

The correct option is B On the surface of sphere $A$
When the spheres are connected by a wire, then potential on both becomes equal
$∴ \frac{Q_{1}}{r_{2}} = \frac{Q_{1}}{r_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}$
when $r_{1} > r_{2}$ then $Q_{1} > Q_{2}$