Question

A particle of charge 'q' and mass 'm' starts moving from the origin under the action of electric field, \( \vec{E}=E_{0}\hat{i} \) with a velocity \( v=v_{0}\hat{j} \). The time taken to increase its velocity to \( \frac{\sqrt{5}}{2} v_0 \) is:

• A. \( \frac{mv_{0}}{qE_{0}} \)
• B. \( \frac{mv_{0}}{2qE_{0}} \)
• C. \( \frac{\sqrt{3}mv_{0}}{2qE_{0}} \)
• D. \( \frac{\sqrt{5}mv_{0}}{2qE_{0}} \)

Hint:

Two-dimensional motion under a perpendicular field is identical to projectile motion. The velocity along the unforced axis stays completely constant, while the velocity along the forced axis grows linearly with time.

Answer 1

The Correct Option is B

Step 1: See the directions clearly.
The electric field points along $x$, so the force and acceleration are along $x$. The starting velocity $v_0$ is along $y$. Since there is no force along $y$, the $y$ part of the velocity never changes.
Step 2: Write the acceleration.
The constant acceleration along $x$ is: \[ a_x = \frac{qE_0}{m} \]
Step 3: Write the velocity parts at time t.
The $x$ part grows with time while the $y$ part stays fixed: \[ v_x = \frac{qE_0}{m}t, \qquad v_y = v_0 \]
Step 4: Build the net speed.
The total speed is the square root of the sum of squares: \[ v_{net} = \sqrt{v_x^{2} + v_y^{2}} \] We want this to equal $\dfrac{\sqrt{5}}{2}v_0$.
Step 5: Square both sides.
\[ \left(\frac{qE_0}{m}t\right)^{2} + v_0^{2} = \frac{5}{4}v_0^{2} \]
Step 6: Solve for the time.
\[ \left(\frac{qE_0}{m}t\right)^{2} = \frac{1}{4}v_0^{2} \;\Rightarrow\; \frac{qE_0}{m}t = \frac{v_0}{2} \] \[ \boxed{t = \frac{mv_0}{2qE_0}} \]