Question 211: Electrostats & Capacitance

Question

A uniform electric field, $$\overrightarrow E  =  - 400\sqrt 3 \widehat y$$ 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$$\sqrt {10} $$ $$ 	imes $$ 106 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 $${q \over m}$$
= 1010 Ckg−1
. Then

Accepted Answer

$$a = {{qE} \over m}{10^{10}} 	imes 400\sqrt 3 \,m/{s^2}$$
or $$a = 4\sqrt 3  	imes {10^{12}}\,m/{s^2}$$
$$ 	herefore $$ $$R = {{{u^2}\sin 2	heta } \over a}$$
$$ \Rightarrow \sin 2	heta  = {{Ra} \over {{u^2}}}$$
or $$\sin 2	heta  = {{5 	imes 4\sqrt 3  	imes {{10}^{12}}} \over {4 	imes 10 	imes {{10}^{12}}}}$$
$$ \Rightarrow \sin 2	heta  = {{\sqrt 3 } \over 2}$$
$$ \Rightarrow 2	heta  = 60^\circ ,\,120^\circ $$
$$ \Rightarrow 	heta  = 30^\circ $$ or $$\,60^\circ $$ for same range
$$ 	herefore $$ $$T = {{2u\sin 	heta } \over a}$$
$$ = {{2 	imes 2\sqrt {10}  	imes {{10}^6}} \over {2 	imes 4\sqrt 3  	imes {{10}^{12}}}}$$
$$ \Rightarrow T = \sqrt {{{10} \over {12}}}  	imes {10^{ - 6}} \Rightarrow T = \sqrt {{5 \over 6}}  	imes {10^{ - 6}}$$

Answer

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

Answer

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

Answer

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

Answer

time taken by the particle to hit T is $$\sqrt {{5 \over 3}} $$ $$\mu $$s

Electrostats & Capacitance

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