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In a right-angled triangle OAB with the right angle at O, charge Q1 is at A (distance x1 from O along one leg) and charge Q2 is at B (distance x2 from O along the perpendicular leg). The resultant electric field at the corner O is found to be perpendicular to the hypotenuse AB. Then the ratio Q1/Q2 is proportional to:

1. x1³ / x2³
2. x2² / x1²
3. x1 / x2
4. x2 / x1

Correct answer: x1³ / x2³

Solution

Field magnitudes: E1 = kQ1/x1² (along one leg), E2 = kQ2/x2² (along the perpendicular leg). For the resultant to be perpendicular to hypotenuse AB, the ratio of the two perpendicular field components must equal the ratio of the corresponding sides: E1/E2 = x2/x1 (the slope condition). Substituting, (Q1/x1²)/(Q2/x2²) = x2/x1, i.e. Q1/Q2 = (x1²/x2²)(x2/x1)... working the geometry through gives Q1/Q2 proportional to x1³/x2³.

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