Question

two identical metal balls with charges +25nC and -5nC separated by some distance,exert a force of magnitude F on each other.they are joined by a conducting wire and then wire is removed . the force between them will now be​

Answer 1

Answer:

O.8F

Explanation:

When they are joined chare flows till both have equal charge

Initial charge = 25-5=20

Final charge = 20/2 = 10 on each

F is proportional to 25×5 =125

New F is proportional to10×10 = 100

New F = 0.8F

Answer 2

The force between them will be now 4F/5.

Two identical metal balls with charges +25nC and -5nC separated by some distance, exert a force of magnitude F on each other. they are joined by a conducting wire and then wire is removed.

We have to find the force between them now.

Concept :

• When two identical balls of different charges are joined by a conducting wire, charge transfers one ball to the other until both have same charge.

Let net charge on each metal ball after wire is removed is Q.

Here, Q = (25 - 5)/2 = 10nC

We assume, distance between the metal balls is r.

Initial Coulomb force,

$F=\frac{\kappa(25nC)\times(5nC)}{r^2} ....(1)$

Final Coulomb force,

$F_f=\frac{\kappa Q^2}{r^2}=\frac{\kappa(10nC)^2}{r^2}...(2)$

From equations (1) and (2) we get,

Final force between them, Ff = 4F/5

Therefore the force between the metal balls will be now 4F/5.

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