A particle of mass $m$ and charge $q$ traverses a distance $d$ from rest in a uniform electric field $E$. Prove that the velocity ($v$) attained by the particle is $v = \sqrt{\dfrac{2qEd}{m}}$.

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> [!TLDR]
> Force $qE$ gives acceleration $qE/m$; kinematics ($v^2=2ad$) then yields $v=\sqrt{2qEd/m}$.
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> **Concept:** A uniform electric field exerts a constant force on the charge, so the motion is uniformly accelerated (like a body in gravity).
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> **Step 1 — Force and acceleration.** The electric force on the particle is $F=qE$. By Newton's second law its acceleration is
> $$a=\frac{F}{m}=\frac{qE}{m}.$$
> **Step 2 — Apply kinematics.** The particle starts from rest ($u=0$) and covers distance $d$. Using $v^2=u^2+2ad$: …

Q.A particle of mass $m$ and charge $q$ traverses a distance $d$ from rest in a uniform electric field $E$ . Prove that the velocity ( $v$ ) attained by the particle is $v = \sqrt{\dfrac{2qEd}{m}}$ .

Concept understanding — Force on a Charge in a Uniform Electric Field

Force on a Charge in a Uniform Electric Field

Q.A particle of mass mmm and charge qqq traverses a distance ddd from rest in a uniform electric field EEE. Prove that the velocity (vvv) attained by the particle is v=2qEdmv = \sqrt{\dfrac{2qEd}{m}}v=m2qEd​​.

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Q.A particle of mass $m$ and charge $q$ traverses a distance $d$ from rest in a uniform electric field $E$ . Prove that the velocity ( $v$ ) attained by the particle is $v = \sqrt{\dfrac{2qEd}{m}}$ .

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