Two point charges Q1 and Q2 exerts force F on each other, when kept at certain distance apart. If the charge on each particle is halved and the distance between them is doubled, then the new force between the two particles would be
A
F2
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B
F4
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C
F8
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D
F16
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Solution
The correct option is DF16 From Coulomb's law:
F=kQ1Q2r2……(1)
Since Q′1=Q12,Q′2=Q22 and r′=2r
From Eq. (1), we get
F′=k(Q12)(Q22)(2r)2
⇒F′=kQ1Q216r2......(2)
from Eq.(1) and Eq.(2), we get
F′=F16
Hence, option (d) is the correct answer.
Why this question ?It intends to test your understanding of direct andinverse proportionality of parameters in Coulomb's law.