JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A particle of charge \(q\) and mass \(m\) is subjected to an electric field \(E = E _{0}\left(1- ax ^{2}\right)\) in the \(x-\)direction, where \(a\) and \(E _{0}\) are constants. Initially the particle was at rest at \(x=0 .\) Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is
- A \(\sqrt{\frac{2}{a}}\)
- B \(\sqrt{\frac{1}{a}}\)
- C \(a\)
- D \(\sqrt{\frac{3}{a}}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\frac{3}{a}}\)
Step-by-step Solution
Detailed explanation
\(E = E _{0}\left(1- ax ^{2}\right)\) \(W =\int q E\, d x = q E _{0} \int_{0}^{ x _{0}}\left(1- ax ^{2}\right) dx\) \(=q E_{0}\left[x_{0}-\frac{a x_{0}^{3}}{3}\right]\) For \(\Delta KE =0, \quad W =0\) Hence \(x _{0}=\sqrt{\frac{3}{ a }}\)
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