Charges Q 1 and Q 2 are at points A and B of a right-angled triangle O A B . The resultant electric field at point O is perpendicular to the hypotenuse, then Q 1 / Q 2 is proportional to:

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Aicharya · Accepted Answer

Using triangle properties tan &#952; = x 1 x 2 tan &#952; = E 2 E 1 = K Q 2 / x 2 2 K Q 1 / x 1 2 &#8658; x 1 x 2 = K Q 2 / x 2 2 K Q 1 / x 1 2 Q 1 Q 2 = x 1 x 2

Charges Q 1 and Q 2 are at points A and B of a right-angled triangle O A B . The resultant electric field at…

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