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:
√2d
Explanation:
f = kq1q2/d^2 = k2q1q2/d'^2
2d^2 = d'^2
d' = √2d
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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.
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