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3 years ago

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Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are se

parated by 12 cm. (Why can you solve this problem without knowing the magnitudes of the charges?)

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

5 N

Explanation:

$F_1$ = 20 N

$r_1$ = 6 cm

$r_2$ = 12 cm

k = Coulomb constat

q = Charge

$\dfrac{F_1}{F_2}=\dfrac{\dfrac{kq_1q_2}{r_1^2}}{\dfrac{kq_2q_2}{r_2^2}}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{12^2}{6^2}\\\Rightarrow \dfrac{F_1}{F_2}=4\\\Rightarrow F_2=\dfrac{F_1}{4}\\\Rightarrow F_2=\dfrac{20}{4}\\\Rightarrow F_2=5\ N$

The force is 5 N

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