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JEE Main 18 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 27
Total: 90
An infinite number of point charges, each carrying
1
µ
C
charge, are placed along the
Y
-axis at
y
=
1
m
,
2
m
,
4
m
,
8
m
. The total force on a
1
C
point charge, placed at the origin, is
x
×
10
3
N
. The value of
x
to the nearest integer, is .............. .
(Take,
1
4
π
ε
0
=
9
×
10
9
N
−
m
2
/
C
2
)
[18 Mar 2021 Shift 2]
Your Answer:
Validate
Solution:
The
1
C
charge present in the origin and other
1
µ
C
charge are placed at
1
m
,
2
m
,
4
m
,
8
m
.
.
.
Using the Coulomb's law,
F
=
k
q
1
q
2
r
2
F
=
9
×
10
9
×
(
1
)
×
10
−
6
[
1
+
1
2
2
+
1
4
2
+
1
8
2
+
⋯
]
=
9
×
10
3
×
[
1
1
−
1
4
]
=
12
×
10
3
N
The value of
x
to the nearest integer is 12 .
© examsnet.com
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