A bullet of mass 2 g carries a charge of \( 2\,\mu\mathrm{C} \). Through what potential difference must it be accelerated, starting from rest, to acquire a speed of \( 10\,\mathrm{m/s} \)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 50\,\mathrm{kV} \)
When a charge \( q \) is accelerated through a potential difference \( V \), the electric field does work \( qV \) on it. Starting from rest, this work becomes its kinetic energy, so \( qV=\frac{1}{2}mv^{2} \).
In SI units the data are:
\( m=2\,\mathrm{g}=2\times10^{-3}\,\mathrm{kg} \)
\( q=2\,\mu\mathrm{C}=2\times10^{-6}\,\mathrm{C} \)
\( v=10\,\mathrm{m/s} \)
The potential difference needed is:
\( \begin{aligned} V &= \frac{mv^{2}}{2q} \\ &= \frac{2\times10^{-3}\,\mathrm{kg}\times(10\,\mathrm{m/s})^{2}}{2\times2\times10^{-6}\,\mathrm{C}} \\ &= \frac{0.2}{4\times10^{-6}}\,\mathrm{V} \\ &= 5\times10^{4}\,\mathrm{V} \\ &= 50\,\mathrm{kV} \end{aligned} \)