Two extremely small charged copper spheres have their centres seperated by a distance of 50 cm in vaccum.what will be the force of repulsion if the chargeon each sphere is doubledand their seperation is halved
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Explanation:
We know that F = q₁q₂/ r²
When charges doubles and separation halved
F'= F 4 ÷(1/4)
= 16F
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