Two particles having mass m and charge q each are to be placed on a rough horizontal surface
with coefficient of static friction . What can be the minimum separation between them so that
they remain stationary?
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Answer:
Answer
Correct option is
C
4πε
0
1
mv
2
4Q
2
Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be u
So using conservation of momentum we get
mv=2mu or u=
2
v
The initial energy of the system is given as
2
1
mv
2
and the energy at the minimum distance is given as
2
1
m(
2
v
)
2
+
2
1
m(
2
v
)
2
+
4πϵ
o
1
R
Q
2
Equating the two energies we get
2
1
mv
2
=
4
1
mv
2
+
4πϵ
o
1
R
Q
2
or
4
1
mv
2
=
4πϵ
o
1
R
Q
2
or
R=
4πϵ
o
mv
2
4Q
2
Explanation:
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