JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2013
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(y\)
Step-by-step Solution
Detailed explanation
\(\Rightarrow \mathrm{F}_{\mathrm{net}}=2 \mathrm{F} \cos \theta\)
\(F_{n e t}=\frac{2 k q\left(\frac{q}{2}\right)}{(\sqrt{y^{2}+a^{2}})^{2}} \cdot \frac{y}{\sqrt{y^{2}+a^{2}}}\)
\(F_{n e t}=\frac{2 k q\left(\frac{q}{2}\right) y}{\left(y^{2}+a^{2}\right)^{3 / 2}} \Rightarrow \frac{k q^{2} y}{a^{3}}\)
So, \(F \propto y\)
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