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

Pick an option to check your answer.
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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