Question

Two identical metallic sphere having unequal opposite charges are are placed

distance of 0.05m apart in air.
After bringing them in contact
with each other, they are again placed at the same distance apart, now the force of repulsion between them is 0.108 N. Calculate the final charge on each of them.​

Answer 1

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 = (9×109 x Q2) / 0.92

Q2 = 0.025 x 0.92 / 9×109

Q = 1.5 x 10-6 C

Step-by-step explanation:

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