JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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
- A \(\frac{ x _{2}^{2}}{ x _{1}^{2}}\)
- B \(\frac{x_{1}^{3}}{x_{2}^{3}}\)
- C \(\frac{x_{1}}{x_{2}}\)
- D \(\frac{ x _{2}}{ x _{1}}\)
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 \(\tan \theta=\frac{ E _{2}}{ E _{1}}=\frac{ x _{1}}{ x _{2}}\)…
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