Question:
A charged particle of mass \(m\) and charge \(q\) is released from rest in an electric field of magnitude \(E\). The kinetic energy of the particle after time \(t\) second is
A charged particle of mass \(m\) and charge \(q\) is released from rest in an electric field of magnitude \(E\). The kinetic energy of the particle after time \(t\) second is
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Start from \(F = qE\), find \(a\), then \(v = at\) (from rest), and \(KE = \frac{1}{2}mv^2\).
Updated On: May 2, 2026
- \(\dfrac{qEt}{2m}\)
- \(\dfrac{q^2E^2t^2}{2m}\)
- \(\dfrac{q^2E^2t^2}{m}\)
- \(\dfrac{q^2E^2t^2}{2m}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Force \(F = qE\), acceleration \(a = qE/m\).
Step 2: Detailed Explanation:
Velocity after time \(t\): \(v = at = \dfrac{qE}{m} \cdot t\)
Kinetic energy: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{qEt}{m}\right)^2 = \frac{q^2E^2t^2}{2m} \]
Step 3: Final Answer:
\(KE = \mathbf{\dfrac{q^2E^2t^2}{2m}}\).
Force \(F = qE\), acceleration \(a = qE/m\).
Step 2: Detailed Explanation:
Velocity after time \(t\): \(v = at = \dfrac{qE}{m} \cdot t\)
Kinetic energy: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{qEt}{m}\right)^2 = \frac{q^2E^2t^2}{2m} \]
Step 3: Final Answer:
\(KE = \mathbf{\dfrac{q^2E^2t^2}{2m}}\).
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