Force of attraction between two charges is $20 N$ . If the distance between them is doubled then the new force of attraction is :

5 N

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20 N

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30 N

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10 N

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Solution

The correct option is C $5 N$
$F \propto \frac{1}{r^{2}}$

Hence, on doubling the distance $r$ , force becomes one fourth.

Hence force becomes $\frac{20}{4} = 5 N$

Answer-(A)

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Force of attraction between two charges is 20N. If the distance between them is doubled then the new force of attraction is :

The correct option is C 5NF∝1r2Hence, on doubling the distance r, force becomes one fourth.Hence force becomes 204=5NAnswer-(A)

Force of attraction between two charges is $20 N$ . If the distance between them is doubled then the new force of attraction is :

The correct option is C $5 N$
$F \propto \frac{1}{r^{2}}$

Hence, on doubling the distance $r$ , force becomes one fourth.

Hence force becomes $\frac{20}{4} = 5 N$

Answer-(A)