Physics, asked by AvinshYadav, 10 months ago

Two conducting spheres, each given a charge q are kept far apart as shown. The amount of charge that crosses
the switch S, when it is closed, is (the connection between the spheres is conducting)

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Answered by BrainlyHero80

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Both the conducting spheres are connected by same wire so they have the same potential. Let x is the charge that crosses from smaller sphere to bigger sphere when switch 1 is closed. Then x amount of charge is subtracted from the total charge in the small sphere and x amount of charge is added in the bigger sphere.

$\frac{k(q-x)}{R}$ = $\frac{k(q-x)}{2 R}$

2q - 2x = q + x

q = 3x

x = $\frac{q}{3}$

$\frac{q}{3}$ amount of charge crosses from smaller sphere to bigger sphere.

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Answered by yashikapandya

Answer:

A parallel plate capacitor is charged and then the battery is disconnected. When the plates of the capacitor are brought closer, then. d is decreased. Hence, C will increase

I hope it's hellpfull

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Two conducting spheres, each given a charge q are kept far apart as shown. The amount of charge that crossesthe switch S, when it is closed, is (the connection between the spheres is conducting)​

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I hope it's hellpfull

Two conducting spheres, each given a charge q are kept far apart as shown. The amount of charge that crosses
the switch S, when it is closed, is (the connection between the spheres is conducting)