Question

# Two charges of equal magnitude and at a distance'r' exert a force $$F$$ on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

A
F/8
B
F/4
C
4F
D
F/16

Solution

## The correct option is D $$F / 16$$$$\textbf{Step 1: Electrostatic force in initial case}$$ Let $$Q_{1} = Q_{2} = Q , d = r$$ By Coulomb's Law:       $$F = \dfrac{KQ^2}{r^2}$$                  $$....(1)$$                                                                                                                                                        $$\textbf{Step 2: Electrostatic force in final case}$$ When charge is halved and separation is doubledSo,     $$Q_{1}' = Q_{2}' = \dfrac{Q}{2} ,\ \ \ d' = 2r$$ So                $$F' = \dfrac{K \left (\dfrac{Q}{2} \right )^2}{(2r)^2} = \dfrac{KQ^2}{16 r^2}$$ From equation $$(1)$$               $$F' = \dfrac{F}{16}$$ Hence, Option $$D$$ is correctPhysics

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