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Electric Potential Due to a Solid Sphere of Charge

Two metal sph...

Question

Two metal spheres of capacitance $C_{1}$ and $C_{2}$ carry some charges. They are put in contact and then separated. The final charge $Q_{1}$ and $Q_{2}$ on them will satisfy

A

\frac{C_{1}}{C_{2}} > \frac{Q_{1}}{Q_{2}}

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B

\frac{C_{1}}{C_{2}} < \frac{Q_{1}}{Q_{2}}

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C

\frac{C_{1}}{C_{2}} = \frac{Q_{1}}{Q_{2}}

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D

\frac{C_{1}}{C_{2}} = \frac{Q_{2}}{Q_{1}}

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Solution

The correct option is C $\frac{C_{1}}{C_{2}} = \frac{Q_{1}}{Q_{2}}$
We know that,

$V = \frac{Q}{C}$

The potential of the two spheres will become equal after they come in contact with each other.

Thus, $V_{1} = V_{2}$

$\Rightarrow \frac{Q_{1}}{C_{1}} = \frac{Q_{2}}{C_{2}}$

$\Rightarrow \frac{C_{1}}{C_{2}} = \frac{Q_{1}}{Q_{2}}$

Therefore, $(c)$ is correct.

Key concept- Two conductors in contact have equal voltage.

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Electric Potential Due to a Solid Sphere of Charge

Standard XII Physics

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