A uniform electric field,vec{E} = -400 sqrt{3 | vedclass

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Question

(D) The acceleration of the particle in the $y$-direction is given by $a_y = \frac{qE_y}{m} = (10^{10})(-400 \sqrt{3}) = -400 \sqrt{3} 	imes 10^{10}

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(D) The acceleration of the particle in the $y$-direction is given by $a_y = \frac{qE_y}{m} = (10^{10})(-400 \sqrt{3}) = -400 \sqrt{3} 	imes 10^{10} 	ext{ ms}^{-2}$.Since the field is in the negative $y$-direction,the particle experiences a downward acceleration. The range $R$ of a projectile is given by $R = \frac{u^2 \sin 2	heta}{|a_y|}$.Given $R = 5 	ext{ m}$ and $u = 2 \sqrt{10} 	imes 10^6 	ext{ ms}^{-1}$,we have $u^2 = 40 	imes 10^{12} 	ext{ m}^2	ext{s}^{-2}$.$5 = \frac{40 	imes 10^{12} \sin 2	heta}{400 \sqrt{3} 	imes 10^{10}} = \frac{4000 \sin 2	heta}{400 \sqrt{3}} = \frac{10 \sin 2	heta}{\sqrt{3}}$.$\sin 2	heta = \frac{5 \sqrt{3}}{10} = \frac{\sqrt{3}}{2}$.Thus,$2	heta = 60^{\circ}$ or $120^{\circ}$,which gives $	heta = 30^{\circ}$ or $60^{\circ}$. So,option $(B)$ is correct.The time of flight is $t = \frac{2u \sin 	heta}{|a_y|}$.For $	heta = 30^{\circ}$,$t_1 = \frac{2 	imes 2 \sqrt{10} 	imes 10^6 	imes (1/2)}{400 \sqrt{3} 	imes 10^{10}} = \frac{2 \sqrt{10}}{400 \sqrt{3}} 	imes 10^{-4} = \frac{\sqrt{10}}{200 \sqrt{3}} 	imes 10^{-4} = \sqrt{\frac{10}{120000}} 	imes 10^{-2} = \sqrt{\frac{1}{12000}} 	imes 10^{-2} = \sqrt{\frac{5}{6}} \mu	ext{s}$.For $	heta = 60^{\circ}$,$t_2 = \frac{2 	imes 2 \sqrt{10} 	imes 10^6 	imes (\sqrt{3}/2)}{400 \sqrt{3} 	imes 10^{10}} = \frac{2 \sqrt{30}}{400 \sqrt{3}} 	imes 10^{-4} = \frac{2 \sqrt{10}}{400} 	imes 10^{-4} = \sqrt{\frac{5}{2}} \mu	ext{s}$.Thus,option $(C)$ is also correct.

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