JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A particle of mass ' \(m\) ' and charge ' \(q\) ' is fastened to one end ' \(A\) ' of a massless string having equilibrium length \(l\), whose other end is fixed at point ' \(O\) '. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x -axis is
- A \(\sqrt{\frac{\mathrm{qE} l}{2 \mathrm{~m}}}\)
- B \(\sqrt{\frac{\mathrm{qE} l}{\mathrm{~m}}}\)
- C \(\sqrt{\frac{\mathrm{qE} l}{4 \mathrm{~m}}}\)
- D \(\sqrt{\frac{2 q \mathrm{El}}{\mathrm{m}}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{\mathrm{qE} l}{\mathrm{~m}}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{W}_{\mathrm{all}}=\Delta \mathrm{k} \\ & \mathrm{W}_{\mathrm{e}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \\ & \mathrm{qE} \frac{\ell}{2}=\frac{1}{2} \mathrm{mv}^2-0 \\ & \mathrm{v}=\sqrt{\frac{\mathrm{qE} \ell}{\mathrm{m}}}\end{aligned}\)
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