Question

A positively charged particle, having charge q, is accelerated by a potential difference V. This particle moving along the x-axis enters a region where an electric field E exists. The direction of the electric field is along positive y-axis. The electric field exists in the region bounded by the line $x=0$ and $x=a$. Beyond the line $x=a$ (i.e., in the region $x>a$), there exists a magnetic field of strength B, directed along the positive y-axis. Find
at which point does the particle meet the line $x=a$

Answer 1

The work done by the potential difference gets stored as its kinetic energy.
$\therefore \frac{1}{2}m{v}^2=qV\Rightarrow v={\left(\frac{2qV}{m}\right)}^{1/2}$ (i)
When it enters the region $xϵ[0,a)$, it experiences an electric field $\overrightarrow{E}=E\hat{j}$
Time taken to cross the region: $vt=a$
$\Rightarrow t=\frac{a}{v}=a{\left(\frac{m}{2qV}\right)}^{1/2}$ (ii)
The distance travelled in y-direction during this time is
$y=\frac{1}{2}\frac{qE}{m}{t}^2=\frac{1}{2}\times \frac{qE}{m}\times a^2\times \frac{m}{2qV}$
$\Rightarrow y=\frac{1}{4}\frac{E{a}^2}{V}$
Hence, the particle meets the line $x=a$ at point
$(x,y)=(a,\frac{1}{4}\frac{E{a}^2}{V})$