Electrostatics Part 3

Electric field of the thread:At distance $r$ from an infinitely long line charge, $E=\frac{\lambda}{2\pi\epsilon_0 r}$ directed radially outward.Circular motion condition:The electron has charge $-e$, so the attractive force toward the wire is $F=\frac{e\lambda}{2\pi\epsilon_0 r}$.This force supplies the required centripetal force:$$\frac{mv^2}{r}=\frac{e\lambda}{2\pi\epsilon_0 r}.$$Cancelling $r$:$$mv^2=\frac{e\lambda}{2\pi\epsilon_0}.$$Thus the kinetic energy is$$K=\frac{1}{2}mv^2=\frac{e\lambda}{4\pi\epsilon_0}.$$Since $e$, $\lambda$, and $\epsilon_0$ are constants, $K$ does not depend on $r$.Therefore the $K$ versus $r$ graph is a horizontal straight line parallel to the $r$-axis.Hence the correct graph is option $3$, i.e. seq $2$.