calculate the forcebetweentwo balls each havinga charge of 12c and are.8 cm Apart
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Answer:
Correct option is
D
13J
Given two charges q
1
=12μC, q
2
=8μC are at distance 'd
1
'=10cm
We have to find the work done in bringing them d
2
=4cm closer.
For this, we find the charge in the potential energy of the system and then take the difference.
Now d
1
=10cm=10×10
−2
=10
−1
m
d
2
=4cm=4×10
−2
m
So, potential (V
1
) when q
1
and q
2
are distance d
1
apart,
V
1
=
4πε
0
1
d
1
q
1
q
2
Potential (V
2
) when q
1
and q
2
are distance d
2
apart,
V
2
=
4πε
0
1
d
2
q
1
q
2
Work done =V
2
−V
1
=
4πε
0
1
q
1
q
2
(
d
2
1
−
d
1
1
)
=9×10
9
×12×8×10
−12
(
4×10
−2
1
−
10×10
−2
1
)
=864×10
−3
[
40×10
−2
6×10
2
]
=129.6×10
1
=12.96≈13Joules
Explanation:
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