Question

A uniform electric field, $ \overrightarrow{E}=-400\sqrt{3\hat{y}} N{C}^{-1}$ is applied in a region. A charged particle of mass $ m$ carrying positive charge $ q$ is projected in this region with an initial speed of $ 2\sqrt{10}\times {10}^{6} m{s}^{-1}$. This particle is aimed to hit a target $ T$, which is $ 5 m$ away from its entry point into the field as shown schematically in the figure. Take $ \frac{q }{ m} = {10}^{10}CK{g}^{-1}$. Then

• A. the particle will hit T if projected at an angle $45°$ from the horizontal
• B. the particle will hit T if projected either at an angle $30°$ or $60°$ from the horizontal
• C. time taken by the particle to hit $T$could be $\sqrt{\frac{5}{2}}\mu s$ as well as $\sqrt{\frac{5}{6}}\mu s$
• D. time taken by the particle to hit $T$is $\sqrt{\frac{5}{3}}\mu s$

Answer 1

Answer: (B), (C)

time taken by the particle to hit $T$could be $\sqrt{\frac{5}{2}}\mu s$ as well as $\sqrt{\frac{5}{6}}\mu s$

Step 1: Given Data:

Uniform Electric field $\overrightarrow{E}=-400\sqrt{3}\hat{y}N{C}^{-1}$

The initial speed of the particle $v=2\sqrt{10}\times {10}^{6}m{s}^{-1}$

Distance of the target from the entry point $R=5m$

The ratio of charge to mass of the particle $\frac{q}{m}={10}^{10}CK{g}^{-1}$

Step 2: Finding the angle at which the particle hit the target $\mathit{T}$:

Let us consider a projectile motion of the particle

If $g_{eff}$ is the effective acceleration then $g_{eff}=\frac{qE}{m}$

$={10}^{10}Ck{g}^{-1}\times \left(400\sqrt{3}N{C}^{-1}\right)$

$=400\sqrt{3}\times {10}^{10}m{s}^{-2}$

Now, if $R$is the horizontal range then $R=\frac{v^2\sin 2\theta }{g_{eff}}$

$\Rightarrow$ $5=\frac{{\left(2\times \sqrt{10}\right)}^2\sin 2\theta }{400\sqrt{3}\times {10}^{10}}$

$\Rightarrow$ $\sin 2\theta =\frac{\sqrt{3}}{2}$

$\Rightarrow$ $\theta =30°,60°$

Step 3. Find the time taken to strike the target $\mathit{T}$:

If $t$ is the time taken to hit the Target $T$ then $t=\frac{2v\sin \theta }{g_{eff}}$

$=\frac{2\times 2\sqrt{10}\times {10}^6\sin 30°}{400\sqrt{3}\times {10}^{10}}$, $\frac{2\times 2\sqrt{10}\times {10}^6\sin 60°}{400\sqrt{3}\times {10}^{10}}$

$=\sqrt{\frac{5}{6}}\mu s$ , $\sqrt{\frac{5}{2}}\mu s$

Hence both options (B) and (C) are correct.