चार्जेस ऑफ 1.2 * 10 के पावर माइनस 8 कुलम एंड 1 पॉइंट 6 * 10 के पावर माइनस साइन टिंकू लमार प्लेस अट टू प्वाइंट्स ए एंड बी डिस्टेंस 5 सेंटीमीटर फ्रॉम ईच अदर कंप्यूट द इलेक्ट्रिक फील्ड अट ए प्वाइंट डिस्टेंस 3 सेंटीमीटर फ्रॉम एंड 4 सेंटीमीटर फ्रॉम बी
if any one complete this numerical I mark him as brainly
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12th
Physics
Electrostatic Potential and Capacitance
Electrostatic Potential Energy and Potential Difference
Two positive point charges ...
PHYSICS
Two positive point charges are of 12C and 8C are 10cm apart from each other. The work done in bringing them 4cm closer is
MEDIUM
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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≈13Joules