Question

two electric charge a distance "d"apart, mutually attract with a force "f".if one of the charges is doubled and to maintain the same force between them, the new separation between the charges must be
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Answer 1

Answer:

√2d

Explanation:

f = kq1q2/d^2 = k2q1q2/d'^2

2d^2 = d'^2

d' = √2d

Answer 2

The final distance = √2 d

Let Q1 and Q2 be the two charges and let d be the distance between the two charges .

Initial Force between the charges =

(K Q1 × Q2)/d²

When one of the charges is doubled (either Q1 or Q2) , the distance between them equals to d'.

Final force = ( 2K × Q1 × Q2)/d'²

Given , initial force = final force

=>(K×Q1 × Q2)/d²= (2K × Q1 × Q2)/d'²

=> d'² = 2d²

=> d' = √2 d

The final distance when one of the charge is doubled = √2 d.