Charge Q1 and Q2 are at points A and B of a right angle triangle OAB (see fig). The resultant electric field at point O is perpendicular to the hypotenuse, then Q1/ Q2 is proportional to:

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x13x23

x22x12

x2 x1

x1 x2

answer is D.

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Detailed Solution

E1sinθ=E2cosθkQ1x12sinθ=kQ2x22cosθQ1Q2=x1x22cotθ=x1x2

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