A bullet of mass $2 \mathrm{~g}$ is having of $2 \mu \mathrm{C}$. Through what potential difference must it…
A bullet of mass $2 \mathrm{~g}$ is having of $2 \mu \mathrm{C}$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of of $10 \mathrm{~m} / \mathrm{s}$ ?
$5 \mathrm{kV}$
$50 \mathrm{kV}$
$5 \mathrm{~V}$
$50 \mathrm{~V}$
Solution
Here we apply $\frac{1}{2} m v^2=q V$
$\begin{gathered}
\Rightarrow V=\frac{1}{2} \times \frac{2 \times 10^{-3} \times 10 \times 10}{2 \times 10^{-6}} \\
=50 \mathrm{kV}
\end{gathered}$