An electron revolves around a very long line charge having density $2 × 10^{-6} Cm^{-1}$. The kinetic energy of the electron is

$2.88 × 10^{-15} J$

The correct answer is Option (3) → $2.88 × 10^{-15} J$

Given: Linear charge density of line $\lambda = 2 \times 10^{-6}~\text{C/m}$

Charge of electron: $e = 1.6 \times 10^{-19}~\text{C}$

Permittivity of free space: $\epsilon_0 = 8.854 \times 10^{-12}~\text{F/m}$

Electric field due to an infinite line charge at distance $r$:

$E = \frac{\lambda}{2 \pi \epsilon_0 r}$

Force on electron:

$F = e E = \frac{e \lambda}{2 \pi \epsilon_0 r}$

Centripetal force required for circular motion:

$F = \frac{m v^2}{r}$

Equating forces:

$\frac{m v^2}{r} = \frac{e \lambda}{2 \pi \epsilon_0 r} \Rightarrow m v^2 = \frac{e \lambda}{2 \pi \epsilon_0}$

Kinetic energy of the electron:

$K = \frac{1}{2} m v^2 = \frac{1}{2} \cdot \frac{e \lambda}{2 \pi \epsilon_0} = \frac{e \lambda}{4 \pi \epsilon_0}$

Substitute numerical values:

$K = \frac{(1.6 \times 10^{-19})(2 \times 10^{-6})}{4 \pi (8.854 \times 10^{-12})}$

$K = \frac{3.2 \times 10^{-25}}{1.112 \times 10^{-10}} \approx 2.88 \times 10^{-15}~\text{J}$

Answer: $K \approx 2.88 \times 10^{-15}~\text{J}$