An electron revolves around a very long line charge having density \( 2\times10^{-6}\,\mathrm{C\,m^{-1}} \). The kinetic energy of the electron is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.88\times10^{-15}\,\mathrm{J} \)
The electric field of a long line charge (Gauss's law) supplies the centripetal force that keeps the electron in a circular path around it.
\( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0\,r} \)
The force on the electron is \( eE \), and it equals the centripetal force:
\( \displaystyle \frac{m v^{2}}{r} = eE = \frac{e\lambda}{2\pi\varepsilon_0\,r} \)
\( \displaystyle m v^{2} = \frac{e\lambda}{2\pi\varepsilon_0} \)
The radius \( r \) has cancelled, so the kinetic energy does not depend on the orbit.
\( \displaystyle K = \frac{1}{2} m v^{2} = \frac{e\lambda}{4\pi\varepsilon_0} \)
\( \displaystyle \frac{1}{4\pi\varepsilon_0} = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( \lambda = 2\times10^{-6}\,\mathrm{C\,m^{-1}} \)
\( \displaystyle K = 9\times10^{9} \times 1.6\times10^{-19} \times 2\times10^{-6} = 2.88\times10^{-15}\,\mathrm{J} \)