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3 years ago

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two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges

1 answer:

Hoochie [10]3 years ago

5 0

The electrostatic force between two charges q1 and q2 is given by
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e =8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km$

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to find out

charges of the beads

solution

we know that force is

force = mass × g

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so we know force for mass

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a).

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