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4 years ago

14

If the distance between two positive point charges is tripled, then the strength of the electrostatic repulsion between them wil

l decrease by a factor of of how much?

1 answer:

vagabundo [1.1K]4 years ago

7 0

Coulomb's Law:
$F=\frac{kq_1q_2}{r^2}$
k is a constant
q1 and q2 are charges
r is the distance

$F_1=\frac{kq_1q_2}{r^2}$

the distance is tripled:
$F_2=\frac{kq_1q_2}{(3r)^2}=\frac{kq_1q_2}{9r^2}$

$\frac{\frac{kq_1q_2}{9r^2}}{\frac{kq_1q_2}{r^2}}= \frac{r^2}{9r^2}=\frac{1}{9}$

The strength of the electrostatic repulsion will decrease by a factor of 9.

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3 years ago

Which equation relates charge, time, and current?

Ierofanga [76]

Answer:

I = Δq / t

Explanation:

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2 years ago

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

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Let the mass of box 1 and box 2 is respectively

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3 years ago

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3 years ago

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B. Exfoliation. Hope I helped you out bro.

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3 years ago