JEE Mains · Physics · STD 12 - 1. Electric charges and fields
Two charges, each equal to \(q\), are kept at \(x = -a\) and \(x = a\) on the \(x-\)axis. A particle of mass \(m\) and charge \(q_0=\frac{q}{2}\) is placed at the origin. If charge \(q_0\) is given a small displacement \((y < < a)\) along the \(y-\)axis, the net force acting on the particle is proportional to
- A \(y\)
- B \(-y\)
- C \(\frac{1}{y}\)
- D \(-\)\(\;\frac{1}{y}\)
Answer & Solution
Correct Answer
(A) \(y\)
Step-by-step Solution
Detailed explanation
\(\Rightarrow \mathrm{F}_{\mathrm{net}}=2 \mathrm{F} \cos \theta\) \(F_{n e t}=\frac{2 k q\left(\frac{q}{2}\right)}{(\sqrt{y^{2}+a^{2}})^{2}} \cdot \frac{y}{\sqrt{y^{2}+a^{2}}}\)…
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