KCET · Physics · Electrostatics
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 qEy
- B \( q E y^{2} \)
- C \( q^{2} E y \)
- D \( q E^{2} y \)
Answer & Solution
Correct Answer
(A) qEy
Step-by-step Solution
Detailed explanation
(A)
Velocity gained after moving a distance \( y \) is \( v^{2}=u^{2}+2 a y \)
Here \( \mathrm{u}=0 \) and \( \mathrm{a}=\frac{q E}{m} \)
\( v^{2}=2\left(\frac{q E}{m}\right) y \)
Kinetic energy, \( E=\frac{1}{2} m v^{2} \)
\( =\frac{1}{2} \times m \times\left(2 \times \frac{q E}{m} \times y\right) \) \( \mathrm{E}=\mathrm{qEy} \)
Velocity gained after moving a distance \( y \) is \( v^{2}=u^{2}+2 a y \)
Here \( \mathrm{u}=0 \) and \( \mathrm{a}=\frac{q E}{m} \)
\( v^{2}=2\left(\frac{q E}{m}\right) y \)
Kinetic energy, \( E=\frac{1}{2} m v^{2} \)
\( =\frac{1}{2} \times m \times\left(2 \times \frac{q E}{m} \times y\right) \) \( \mathrm{E}=\mathrm{qEy} \)
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