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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 2 Questions
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© examsnet.com
Question : 17
Total: 100
A point charge
2
×
10
−
2
C
is moved from
P
to
S
in a uniform electric field of
30
NC
−
1
directed along positive
x
-axis. If coordinates of
P
and
S
are
(
1
,
2
,
0
)
m
and
(
0
,
0
,
0
)
m
respectively, the work done by electric field will be
[29-Jan-2023 Shift 2]
1200
mJ
600
mJ
−
600
mJ
−
1200
mJ
Validate
Solution:
ω
E
=
q
→
E
⋅
→
S
=
2
×
10
−
2
[
30
^
i
⋅
(
−
^
i
)
]
=
2
×
10
−
2
(
−
30
)
=
−
60
×
10
−
2
=
−
60
100
=
−
0.6
J
=
−
600
mJ
© examsnet.com
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