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A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$ travelling with a velocity $v$ along the $\mathrm{x}$-axis enters a uniform electric field $\overrightarrow{\mathrm{E}}$ directed along the y-axis. What will be the trajectory of the particle?
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Verified Answer
The correct answer is:
Parabolic
$$
\begin{aligned}
& \text { } \mathrm{y}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \quad(\because \mathrm{u}=0) \\
& \mathrm{y}=\frac{1}{2} \mathrm{at}^2 \\
& \mathrm{~F}=\mathrm{ma}=\mathrm{qE} \\
& \mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}} \\
& \mathrm{y}=\frac{1}{2} \frac{\mathrm{qE}}{\mathrm{m}} \frac{\mathrm{x}^2}{\mathrm{v}^2} \quad(\because \mathrm{x}=\mathrm{vv}) \\
& \mathrm{x}^2=\frac{2 \mathrm{mv}}{\mathrm{qE}} \mathrm{y}
\end{aligned}
$$
Compare with $\mathrm{x}^2=4$ ay
So, the trajectory of the particle will be parabolic.
\begin{aligned}
& \text { } \mathrm{y}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \quad(\because \mathrm{u}=0) \\
& \mathrm{y}=\frac{1}{2} \mathrm{at}^2 \\
& \mathrm{~F}=\mathrm{ma}=\mathrm{qE} \\
& \mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}} \\
& \mathrm{y}=\frac{1}{2} \frac{\mathrm{qE}}{\mathrm{m}} \frac{\mathrm{x}^2}{\mathrm{v}^2} \quad(\because \mathrm{x}=\mathrm{vv}) \\
& \mathrm{x}^2=\frac{2 \mathrm{mv}}{\mathrm{qE}} \mathrm{y}
\end{aligned}
$$
Compare with $\mathrm{x}^2=4$ ay
So, the trajectory of the particle will be parabolic.
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