Physics: The force between a pair of 0.001 c charges is 200 n. what is the distance between them?

The force between a pair of 0.001 c charges is

Ответ:

lilbloke

31.12.2021

From Columb's Law which describe the electrostatic interaction.
The formula is given and looks as follows $F=k \frac{|q1*q2|}{r^{2} }$
where:
F is force acting between charges,
q1 - charge one given in columbs,
q2 - charge two given in columbs,
r - distance between charge in meters
k - constant

First, we have to convert out formula to get r.
$F=k \frac{q1*q2}{r^{2} }$
Now we multiplaty by $r^{2}$ both sides of equation
$F* r^{2} =k*|q1*q2|$
Now we devide both sides by F
$r^{2} =k \frac{|q1*q2|}{F}$
Next step is take square root
$r= \sqrt{k \frac{|q1*q2|}{F}}$
We have all date, we can subsitute it to our formula
$r= \sqrt{k* \frac{|q1q2"}{F} } = \sqrt[2]{8.9* 10^{9* \frac{0.001*0.001}{200} }} = 6.67 [m]$

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Ответ:

cameron1357

17.12.2020

Physics

i need points

Explanation:

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The force between a pair of 0.001 c charges is 200 n. what is the distance between them?

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