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An electron with an initial speed of $4.0 \times 10^{6} \mathrm{~ms}^{-1}$ is brought to rest by an electric field. The mass and charge of an electron are $9 \times 10^{-31} \mathrm{~kg}$ and $1.6 \times 10^{-19} \mathrm{C}$, respectively. Identify the correct statement
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The correct answer is:
The electron moves from a region of higher potential to lower potential through a potential difference of $45\mathrm{~V}$.
$\begin{array}{l}
q V=\frac{1}{2} m v^{2} \\
V=\frac{1}{2} \frac{m v^{2}}{q} \\
V=\frac{1}{2} \times \frac{9 \times 10^{-31} \times\left(4 \times 10^{6}\right)^{2}}{1.6 \times 10^{-19}}=45 V
\end{array}$
$45 \mathrm{~V}$ from higher to lower potential.
q V=\frac{1}{2} m v^{2} \\
V=\frac{1}{2} \frac{m v^{2}}{q} \\
V=\frac{1}{2} \times \frac{9 \times 10^{-31} \times\left(4 \times 10^{6}\right)^{2}}{1.6 \times 10^{-19}}=45 V
\end{array}$
$45 \mathrm{~V}$ from higher to lower potential.
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