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 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:
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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.
Updated On: Jun 8, 2026
- \( \frac{mv_{0}}{qE_{0}} \)
- \( \frac{mv_{0}}{2qE_{0}} \)
- \( \frac{\sqrt{3}mv_{0}}{2qE_{0}} \)
- \( \frac{\sqrt{5}mv_{0}}{2qE_{0}} \)
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The Correct Option is B
Solution and Explanation
Concept:
The electric field acts purely along the x-axis (\( \vec{E} = E_0\hat{i} \)), creating a constant acceleration \( a_x = \frac{qE_0}{m} \) in that direction. The initial velocity is completely along the y-axis (\( v_y = v_0 \)) and remains unchanged over time since there is no force acting along the y-axis.
The net speed at any time \( t \) is given by the Pythagorean combination of components:
\[
v_{\text{net}} = \sqrt{v_x^2 + v_y^2}
\]
Step 1: Expressing the velocity components at time \( t \).
• Horizontal component: \( v_x = a_x \cdot t = \frac{qE_0}{m}t \)
• Vertical component: \( v_y = v_0 \)
Step 2: Equating the net velocity to the target value.
We want the net speed to equal \( \frac{\sqrt{5}}{2}v_0 \): \[ \sqrt{\left(\frac{qE_0}{m}t\right)^2 + v_0^2} = \frac{\sqrt{5}}{2}v_0 \] Squaring both sides of the equation to clear the radical: \[ \left(\frac{qE_0}{m}t\right)^2 + v_0^2 = \frac{5}{4}v_0^2 \]
Step 3: Isolating time parameter \( t \).
\[ \left(\frac{qE_0}{m}t\right)^2 = \frac{5}{4}v_0^2 - v_0^2 = \frac{1}{4}v_0^2 \] Taking the square root on both sides: \[ \frac{qE_0}{m}t = \frac{1}{2}v_0 \implies t = \frac{mv_0}{2qE_0} \] This matches option (B) perfectly.
Step 1: Expressing the velocity components at time \( t \).
• Horizontal component: \( v_x = a_x \cdot t = \frac{qE_0}{m}t \)
• Vertical component: \( v_y = v_0 \)
Step 2: Equating the net velocity to the target value.
We want the net speed to equal \( \frac{\sqrt{5}}{2}v_0 \): \[ \sqrt{\left(\frac{qE_0}{m}t\right)^2 + v_0^2} = \frac{\sqrt{5}}{2}v_0 \] Squaring both sides of the equation to clear the radical: \[ \left(\frac{qE_0}{m}t\right)^2 + v_0^2 = \frac{5}{4}v_0^2 \]
Step 3: Isolating time parameter \( t \).
\[ \left(\frac{qE_0}{m}t\right)^2 = \frac{5}{4}v_0^2 - v_0^2 = \frac{1}{4}v_0^2 \] Taking the square root on both sides: \[ \frac{qE_0}{m}t = \frac{1}{2}v_0 \implies t = \frac{mv_0}{2qE_0} \] This matches option (B) perfectly.
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