Question

A uniform electric field, E=-400$\sqrt3\vec{Y}$YNC-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√10 × 106 ms−1. This particle is aimed to hit a target T, which is 5 m away from its entry point into the field schematically in the figure. Take $\frac{q}{m}$ = 1010 ckg-1.

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

Answer 1

Answer: (B), (C)

the particle will hit T if projected either at an angle of 30º or 60º from the horizontal

Answer 2

The range 𝑅 is given by $R = \frac{2u \sin \theta}{g_{\text{eff}}} \times u \cos \theta = 5 \, \text{m}$
We find $\sin(2\theta) = \frac{qER}{\mu^2} = \frac{\sqrt{3}}{2}$
So, $\theta = 30^\circ$
Therefore, for the same range, the angle of projection could be either $30^∘$ or $60^∘$.
Thus, Option (B) is correct.

The time of flight 𝑇 is given by $T = \frac{2u \sin \theta}{g_{\text{eff}}}$
$=\left(\sqrt{\frac{10}{3}} \times 10^{-6}\right) \sin \theta$
$T_1 \text{ at } \theta = 30^\circ \text{ is } \frac{\sqrt{5}}{6}\mu s$
$T_2 \text{ at } \theta = 60^\circ \text{ is } \frac{\sqrt{5}}{2}\mu s$
So, Option (C) is also correct.