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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2020
  • Hard
  • MCQ

Charges \(Q _{1}\) and \(Q _{2}\) arc at points \(A\) and \(B\) of a right angle triangle \(OAB\) (see figure). The resultant electric field at point \(O\) is perpendicular to the hypotenuse, then \(Q _{1} / Q _{2}\) is proportional to

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Answer & Solution

Correct Answer

(C) \(\frac{x_{1}}{x_{2}}\)

Step-by-step Solution

Detailed explanation

\(E _{2}=\) electric field due to \(Q _{2}\)

\(=\frac{ kQ _{2}}{ x _{2}^{2}}\)

\(E _{1}=\frac{ k Q _{1}}{ x _{1}^{2}}\)

From diagram

\(\tan \theta=\frac{ E _{2}}{ E _{1}}=\frac{ x _{1}}{ x _{2}}\)

\(\frac{ kQ _{2}}{ x _{2}^{2} \times \frac{ k Q _{1}}{ x _{1}^{2}}}=\frac{ x _{1}}{ x _{2}}\)

\(\frac{ Q _{2} x _{1}^{2}}{ Q _{1} x _{2}^{2}}=\frac{ x _{1}}{ x _{2}}\)

\(\frac{Q_{2}}{Q_{1}}=\frac{x_{2}}{x_{1}}\)

\(\frac{Q_{1}}{Q_{2}}=\frac{x_{1}}{x_{2}}\)

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