Question

# F is the force between two charges. If the distance between them is tripled, then the force between the charges will be :

A
F
B
F3
C
F9
D
F27

Solution

## The correct option is B $$\dfrac{F}{9}$$$$F=\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{r^2}$$ and$$F'=\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{(3r)^2}$$$$=\dfrac{1}{9}\Bigg(\dfrac{1}{4\pi \epsilon_0}\,\dfrac{q_1q_2}{r^2}\Bigg)=\dfrac{F}{9}$$Physics

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