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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 2 Questions
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Question : 16
Total: 100
A point charge
q
1
=
4
q
0
is placed at origin. Another point charge
q
2
=
−
q
0
is placed at
x
=
12
cm
. Charge of proton is
q
0
. The proton is placed on
x
-axis so that the electrostatic force on the proton in zero. In this situation,the position of the proton from the origin is _______
cm
.
[29-Jan-2023 Shift 1]
Your Answer:
Validate
Solution:
q
0
x
2
=
4
q
0
(
x
+
12
)
2
x
+
12
=
2
x
x
=
12
Distance from origin
=
x
+
12
=
24
cm
.
© examsnet.com
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