A pair of equal charges are separated by .75 m. There's a 2500 N force acting on them. What is the strength of the charge? Plz explain how you got the answer too.

Answers

Answer 1

Answer:

The magnitude of each charge is 3.94 x $10^(-4)$ coulombs.

We have a pair of equal charges are separated by 0.75 m and there's a 2500 N force acting on them.

We have to strength of the charge.

What is the formula to calculate the force between two point charges separated by a some distance?

According to Coulomb's law, the formula to calculate the force between the two point charges (say q and Q) separated by a some distance (say r) is given by -

F = $(1)/(4\pi \epsilon_(o) ) (qQ)/(r^(2) )$

According to the question -

Force between two point charges equals to = F = 2500 N

Distance between the point charges = 0.75 m

Let's assume the magnitude of each charge be x, therefore using the coulomb's law, we get -

F = $(1)/(4\pi \epsilon_(o) ) (qQ)/(r^(2) )$

$x^(2) = (2500 * 0.75* 0.75)/(9* 10^(9) )\nx^(2) = (2500* 75* 75)/(9* 10^(13) ) \nx^(2) = 1.56* 10^(-7)$

$x = 0.000394\nx = 3.94 * 10^(-4)$ coulombs

Hence, the magnitude of each charge is 3.94 x $10^(-4)$ coulombs.

To understand more about coulomb's law visit the link below -

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Answer 2

Answer:

Answer:

4.5643546 × 10^ -4

Explanation:

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Answers

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Answers

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Answers

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