Question

A charged particle having charge $q$ and mass $m$ is projected into a region of uniform electric field of strength $\overrightarrow{E_0}$, with velocity $\overrightarrow{V_0}$ perpendicular to $\overrightarrow{E_0}$. Throughout the motion, apart from electric force, the particle also experiences a dissipative force of constant magnitude $q{E}_0$ and directed opposite to its velocity. If $\overrightarrow{V_0}|=6\text{m/s}$, then find its speed when it has turned through an angle of ${90}^{\circ }$.

Write speed in m/s.

Answer 1

Here $v$ is speed which is equal to
$v=\sqrt{v_x^2+v_y^2}$
and $\frac{dv}{dt}=$ rate of change in speed
Consider the situation when the particle has turned through an angle of $\psi$
$\frac{dv}{dt}=-\frac{q{E}_0}{m}[1-\sin \psi ]$
$\frac{d{v}_x}{dt}=\frac{q{E}_0}{m}[1-\sin \psi ]$
$\Rightarrow \frac{dv}{dt}=-\frac{d{v}_x}{dt}$
$\Rightarrow v=-v_x+C$
At $t=0,v=v_0\text{and}{v}_x=0$
$\Rightarrow C=v_0$
$\Rightarrow v=v_0-v_x=v_0-v\sin \varphi$
When it has turned through an angle of ${90}^{\circ }$,
$v=\frac{v_0}{1+\sin \varphi }=\frac{6}{1+1}=3\text{m/s}$