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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 2 Questions
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© examsnet.com
Question : 60
Total: 100
A positive charge particle of
100
mg
is thrown in opposite direction to a uniform electric field of strength
1
×
10
5
NC
−
1
. If the charge on the particle is
40
µ
C
and the initial velocity is
200
ms
−
1
, how much distance it will travel before coming to the rest momentarily :
[29-Jun-2022-Shift-1]
1
m
5
m
10
m
0.5
m
Validate
Solution:
v
2
−
u
2
=
2
a
s
⇒
0
2
−
200
2
=
2
(
−
q
E
m
)
(
S
)
⇒
−
200
2
=
2
[
−
40
×
10
−
6
×
10
5
100
×
10
−
6
]
[
S
]
⇒
S
=
4
2
×
4
m
=
0.5
m
© examsnet.com
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