Question

A particle of mass $m$ and positive charge $q$ is projected towards an infinitely long line of a charge (having linear density of charge $+\lambda$) from a distance $t_0$. The direction of initial velocity $v_0$ makes an angle ${30}^{\circ }$ with the normal to the line of charge as shown in figure. The minimum distance of approach of the charge particle with the line of charge will be (neglect gravity). Take $\lambda =\frac{\pi {\epsilon }_{0}m{v}_0^2}{4q}$.

• A. $\frac{r_0}{e}$
• B. $\frac{r_0}{e^2}$
• C. $\frac{r_0}{e^3}$
• D. $\frac{r_0}{2}$

Answer 1

The correct option is C $\frac{r_0}{e^3}$

Particle of mass $=m$ positive charge $=\varphi$

$\lambda =\frac{\pi {ϵ}_{0}m{V}_0^2}{4q};\theta ={30}^0$

distance $=t_0$

initial velocity $=V_0$

the $d=\frac{r_0}{e^3}\cos {30}^0$

$=\frac{r_0}{e^3}.\frac{1}{2}$ $\bullet \frac{1}{2}$

$2d=\frac{r_0}{e^3}$