3 years ago

I need help in this please as soon as possible

Physics

1 answer:

topjm [15]3 years ago

7 0

Answer:

u itls

Explanation:

sorry look it up

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In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

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Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

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Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

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Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

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$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

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