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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
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B
F3
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C
F9
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D
F27
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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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