Question

Q. A particle of mass \(m\) and charge \(q\) is placed at rest in uniform electric field \(E\) and then released. The kinetic energy attained by the particle after moving a distance \(y\) is

(A) \(q E y\)
B) \(q E y^{2}\)
C \(q^{2} E y\)
D \(q E^{2} y\)

Answer 1

Solution:
Velocity gained after moving a distance \(\mathrm{y}\) is \(v^{2}=u^{2}+2 a y\)
Here \(u=0\) and \(a=\frac{q E}{m}\)
\(v^{2}=2\left(\frac{q E}{m}\right) y\)
Kinetic energy, \(E=\frac{1}{2} m v^{2}\)
\(=\frac{1}{2} \times m \times\left(2 \times \frac{q E}{m} \times y\right)\) \(E=q E y\)

Answer 2

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Answer 3

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