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- 24cm from+9q
- 12cm from+9q
- 24cm from+q
- 12cm from+q

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Solution

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Two point charges +9q and +9.

Kept apart =16cm=16×10−2m

Two cases will be needed.

Given charges, +q, +9q. let, the third charge q be placed at a distance x from q and hence its distance from 9q and (16−x)cm for the system to remain in equilibrium F1=F2

Therefore,

F1=K(q×Q)x2

F2=K(9q×Q)(16−x)2

Equating F1 and F2 we get,

9x2=1(16−x)2

F1=F2

or, K(q×Q)x2=K(9q×Q)(16−x)2

or, 1x2=9(16−x)2

or, 1x=9(16−x)

or, 16−x=3x

or, 4x=16

or, x=4

∴ (16−4)cm=12cm from +9q.

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