Question:
The velocity acquired by a charged particle of mass \(m\) and charge \(Q\) accelerated from rest by a potential \(V\) is
The velocity acquired by a charged particle of mass \(m\) and charge \(Q\) accelerated from rest by a potential \(V\) is
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Always equate electric potential energy with kinetic energy for acceleration problems.
Updated On: May 8, 2026
- \(\frac{QV}{m}\)
- \(\sqrt{\frac{m}{QV}}\)
- \(\sqrt{mQV}\)
- \(mQV\)
- \(\sqrt{\frac{2QV}{m}}\)
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The Correct Option is
Solution and Explanation
Concept:
When a charged particle is accelerated through a potential difference, electrical potential energy converts into kinetic energy:
\[
QV = \frac{1}{2}mv^2
\]
Step 1: Write energy conservation. \[ \text{Gain in KE} = \text{Loss in PE} \Rightarrow QV = \frac{1}{2}mv^2 \]
Step 2: Solve for velocity. \[ v^2 = \frac{2QV}{m} \]
Step 3: Take square root. \[ v = \sqrt{\frac{2QV}{m}} \]
Step 4: Conclusion. \[ \boxed{\sqrt{\frac{2QV}{m}}} \]
Step 1: Write energy conservation. \[ \text{Gain in KE} = \text{Loss in PE} \Rightarrow QV = \frac{1}{2}mv^2 \]
Step 2: Solve for velocity. \[ v^2 = \frac{2QV}{m} \]
Step 3: Take square root. \[ v = \sqrt{\frac{2QV}{m}} \]
Step 4: Conclusion. \[ \boxed{\sqrt{\frac{2QV}{m}}} \]
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