A uniform electric field, is applied in a region. A charged particle of mass carrying positive charge is projected in this region with an initial speed of This particle is aimed to hit a target , which is 5 m away from its entry point into the field a

Question

A uniform electric field, $\vec{E} = - 400 \sqrt{3} \hat{y} {NC}^{- 1}$ 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 $2 \sqrt{10} \times 10^{6} {ms}^{- 1} .$ 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 $\frac{q}{m} = 10^{10} C {kg}^{- 1} .$ Then [2020]

Smart Achievers · Accepted Answer

(2, 3)

$Given: electric field, {\vec{E}}_{y} = - 400 \sqrt{3} {N C}^{- 1}$

$Initial speed of charged particle, u = 2 \sqrt{10} \times 10^{6} m/s$

$Range, R = 5 m$

$\frac{q}{m} = 10^{10} C {kg}^{- 1} .$ $∴$ $a_{y} = \frac{q}{m} E_{y} = - 400 \sqrt{3} \times 10^{10}$ $[∵ \frac{q}{m} = 10^{10} {C kg}^{- 1} (given)]$

$Range, R = \frac{u^{2} \sin 2 θ}{a_{y}}$ $\Rightarrow 5 = \frac{40 \times 10^{12} \sin 2 θ}{400 \sqrt{3} \times 10^{10}}$

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

$\Rightarrow 2 θ = 60 ° or 120 °$ $\Rightarrow θ = 30 ° or 60 °$

$Particle hits the target if θ = 30 ° or θ = 60 °$

$u = 2 \sqrt{10} \times 10^{6} {m s}^{- 1}$

$Time of flight$ $T_{1} = \frac{2 u \sin θ}{a_{y}} = \frac{2 \times 2 \sqrt{10} \times 10^{6} \times \frac{1}{2}}{400 \sqrt{3} \times 10^{10}}$

$= \sqrt{\frac{5}{6}} μ s (for θ = 30 °)$

$Time of flight$ $T_{2} = \frac{2 u \sin θ}{a_{y}} = \frac{2 \times 2 \sqrt{10} \times 10^{6} \times \frac{\sqrt{3}}{2}}{400 \sqrt{3} \times 10^{10}}$

$= \sqrt{\frac{5}{6}} μ s (for θ = 60 °)$

Solution

1 the particle will hit $T$ if projected at an angle $45 °$ from the horizontal

2 the particle will hit $T$ if projected either at an angle $30 °$ or $60 °$ from the horizontal

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

4 time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} μ s$

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Solution

1 the particle will hit T if projected at an angle 45° from the horizontal

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

3 time taken by the particle to hit T could be 56μs as well as 52μs

4 time taken by the particle to hit T is 53μs

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1 the particle will hit $T$ if projected at an angle $45 °$ from the horizontal

2 the particle will hit $T$ if projected either at an angle $30 °$ or $60 °$ from the horizontal

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

4 time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} μ s$