A charged particle moving with a certain velocity enters a uniform electric field perpendicular to its direction. In this context, which of the following statements are correct?
(A) Force acting on the particle is constant
(B) Acceleration of the particle is constant
(C) Velocity of the particle is constant
(D) Displacement of the particle follows a parabolic path
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
Take the initial velocity \( v \) along \( x \) and the uniform field \( E \) along \( y \). The force on the charge is \( qE \), and it acts along \( y \), at right angles to the initial velocity.
(A) Force acting on the particle is constant — Correct. The field is uniform, so \( F = qE \) has the same size and direction everywhere.
(B) Acceleration of the particle is constant — Correct. With a constant force and a constant mass, the acceleration is constant.
\( \displaystyle a = \frac{qE}{m} \)
(C) Velocity of the particle is constant — Incorrect. The \( y \)-component of velocity keeps growing (\( v_y = at \)), so both the speed and the direction change.
(D) Displacement of the particle follows a parabolic path — Correct. Along \( x \) the motion is uniform and along \( y \) it is uniformly accelerated, as for a projectile. Eliminating \( t \) gives a parabola.
\( \displaystyle x = vt, \qquad y = \frac{1}{2}at^{2} \;\Rightarrow\; y = \frac{a}{2v^{2}}\,x^{2} \)