If the distance between two stationary charged particle is doubled, the magnitude of the electro-static force between them will be
doubled
halved
a quarter of its former value
unchanged
four times the original value
Correct answer is C
F \(\alpha\) \(\frac{1}{d^2}\)
F x d2 = constant
F1 x (d1)2 = F2 x (d2)2
d2 = 2d1
F1 x (d1)2 = F2 x (2d1)2
F2 = \(\frac{1}{4}\)F1