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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
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B
F/4
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C
4F
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D
F/16
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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 doubled
So,     $$ Q_{1}' = Q_{2}' = \dfrac{Q}{2} ,\ \ \  d' = 2r $$ 
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 correct

Physics

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