Question

At a distance $L$ from a uniformly long wire, a charged particle is thrown radially outwards with a velocity $u$ in the direction perpendicular to the wire. When the particle reaches a distance $2L$ from the wire, its speed is found to be $u\sqrt{2}$ . The magnitude of the velocity when it is at a distance of $16L$ away from the wire is $pu$, where $p$ is (ignore gravity)

Answer 1

Let $V_A,V_B,V_C$ be velocity of particle at points $A,B,C$

Potential difference of any point $A$ wrt point $B$ due to infinite line charge is $\Delta V=\frac{\lambda }{2\pi {ϵ}_0}ln\left(\frac{r_b}{r_a}\right)$
$\Rightarrow V_A-V_B=\frac{\lambda }{2\pi {ϵ}_0}ln\left(2\right)$
Potential difference between $A$ and $C$ is
$\Rightarrow V_A-V_C=\frac{\lambda }{2\pi {ϵ}_0}ln\left(16\right)$

From conservation of energy between A and B
$q(V_A-V_B)=\frac{1}{2}m{V}_B^2-\frac{1}{2}m{V}_A^2$
$\Rightarrow \frac{q\lambda }{2\pi {ϵ}_0}ln\left(2\right)=\frac{1}{2}m(2u^2-u^2)=\frac{1}{2}m{u}^2.......\left(1\right)$
From conservation of energy between A and C
$q(V_A-V_C)=\frac{1}{2}m{V}_c^2-\frac{1}{2}m{u}^2$
$\frac{q\lambda }{2\pi {ϵ}_0}ln\left(16\right)=\frac{1}{2}m{V}_c^2-\frac{1}{2}m{u}^2$
Substituting from $\left(1\right)$
$4\left[\frac{1}{2}m{u}^2\right]=\frac{1}{2}m{V}_C^2-\frac{1}{2}m{u}^2$
$\Rightarrow V_C=u\sqrt{5}$