Two positive charges of 12μC and 8μC are 10 cm apart. The work is done in bringing them 4cm closer so that they are 6cm apart
choose
7.56 J
5.76 J
57.6 J
Zero
Answers
Answer:
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≈13JoulesGiven 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:
Answer:
option c is the correct answer
Explanation:
First chargeq_{1}=12\times10^{-6}\ Cq
1
=12×10
−6
C
Second chargeq_{2}=8\times10^{-6}\ Cq
2
=8×10
−6
C
The distance r_{1}= 10\ cm=0.1\ mr
1
=10 cm=0.1 m
The distance r_{2}= 6\ cm=0.6\ mr
2
=6 cm=0.6 m
so
W= kq_{1}q_{2}(\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}})W=kq
1
q
2
(
r
2
1
−
r
1
1
)
W=9\times10^{9}\times12\times10^{-6}\times8\times10^{-6}\times(\dfrac{1}{0.06}-\dfrac{1}{0.1})W=9×10
9
×12×10
−6
×8×10
−6
×(
0.06
1
−
0.1
1
)
W= 5.76\ JW=5.76 J