KCET · Physics · Electrostatics
A charged particle of mass \(m\) and charge \(q\) is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after \(t\) seconds is
- A \(\frac{E q^2 m}{2 t^2}\)
- B \(\frac{E q m}{t}\)
- C \(\frac{E^2 q^2 t^2}{2 m}\)
- D \(\frac{2 E^2 t^2}{m q}\)
Answer & Solution
Correct Answer
(C) \(\frac{E^2 q^2 t^2}{2 m}\)
Step-by-step Solution
Detailed explanation
According to given situation, force on charge particle in uniform electric field, \(F=q E\)
Acceleration of charge particle,
\(a=\frac{F}{m}=\frac{q E}{m}\)
Velocity of charge particle after time \(t\) is given as
\(v=u+a t=0+\frac{q E}{m} t \Rightarrow v=\frac{q E t}{m}\)
\(\therefore\) Kinetic energy of the charged particle
\(K=\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{q E t}{m}\right)^2=\frac{E^2 q^2 t^2}{2 m}\)
Acceleration of charge particle,
\(a=\frac{F}{m}=\frac{q E}{m}\)
Velocity of charge particle after time \(t\) is given as
\(v=u+a t=0+\frac{q E}{m} t \Rightarrow v=\frac{q E t}{m}\)
\(\therefore\) Kinetic energy of the charged particle
\(K=\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{q E t}{m}\right)^2=\frac{E^2 q^2 t^2}{2 m}\)
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