Force between two identical metal spheres A and
B with charges as shown is F. A third identical, but
uncharged sphere C is brought in contact first with
A, then with B and finally placed at the mid-point
of the line joining A and B. The force on sphere C is
(1) F
(2) 1.5 F
(3) 1.33 F
(4) 0.75 F
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Answer:
Let the charge on each sphere be q then the force on each sphere is F=
r
2
kqq
=
r
2
kq
2
=2×10
−5
.....(1)
When sphere C is touched by sphere A, charge q gets equally divided between the two.
Thus, F
CA
=
r
2
/4
k(q/2)(q/2)
=
r
2
kq
2
and F
CB
=
r
2
/4
k(q/2)(q)
=
r
2
2kq
2
Net force =F
CB
−F
CA
=
r
2
2kq
2
−
r
2
kq
2
=
r
2
kq
2
=2×10
−5
N (using (1))
As F
CB
>F
CA
so net force is away sphere
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