Question

Two spheres have equal and opposite charges and are at a distance of 90 cm from each other. They are touched together and then are placed at their initial position; then they show a repulsive force of 0.025 N. The final charge on them will be :
(a) 1.5 μC
(b) 1.5 C
(c) 3 C
(d) 5 C

Answer 1

Answer:

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Answer 2

Answer:

Let the initial charges be q1 and q2 respectively.

After they come in contact, the charges are rearranged such that they acquire same charge.

let us say that charge on each of them is Q.

They are again brought apart at a distance of 0.9 m. Hence, the force between them will be given as

F = kQ2 / r2

0.025 = (9x109 x Q2) / 0.92

Q2 = 0.025 x 0.92 / 9