determine the force of interaction between the charges -30micro coulomb and +20micro coulomb separated by 10 mm apart
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determine the force of interaction between the charges -30micro coulomb and +20micro coulomb separated by 10 mm apart
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Answer:
Redistribution of charges takes place.
Charge q
1
=3μC
Charge q
2
=8μC
When third charge q
3
=−5μC is added to each, then new charges on q
1
and q
2
will be
q
1
′
=3−5=−2μC
and, q
2
′
=8−5=3μC
Now,
In first case,
40=
4πε
0
1
⋅
r
2
3×8
In second case,
F=
4πε
0
1
×
r
2
(−2×3)
∴
40
F
=
3×8
−2×3
or F=−10N