Question 829: Electrostatics & 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 $ 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 ${q \over m}$
= 10
10
Ckg
−1
. Then
![](https://imagex.cdn.examgoal.net/jgy2xukfm38cuw/b4bdece1-27f7-4271-92dd-145d90921dde/f514c980-014a-11eb-8293-3b6cc591631c/file-jgy2xukfm38cux-360w.png)

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}}$
![](https://imagex.cdn.examgoal.net/1ks6fslqb/b7984daa-d80e-4893-ab18-e46f4d4601fb/5b33ce20-fa0e-11eb-b320-cf8b1e7f97f3/file-1ks6fslqc.jpeg)

Answer

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

Answer

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

Electrostatics & Capacitance

Choose the correct answer: