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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if the distance between the particles is decreased to 1?

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Answer: 539.4 N

Step-by-step explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force
F_(E) between two point charges
q_(1) and
q_(2) is proportional to the product of the charges and inversely proportional to the square of the distance
d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:


F_(E)= K(q_(1).q_(2))/(d^(2)) (1)

Where:


F_(E)=60 N is the electrostatic force


K=8.99(10)^(9) Nm^(2)/C^(2) is the Coulomb's constant


q_(1) and
q_(2) are the electric charges


d=3 m is the separation distance between the charges

Then:


60 N= 8.99(10)^(9) Nm^(2)/C^(2)(q_(1).q_(2))/((3 m)^(2)) (2)

Isolating
q_(1) and
q_(2):


q_(1)q_(2)=6(10)^(-8) C^(2) (3)

Now, if we keep the same charges but we decrease the distance to
d_(1)=1 m, (1) is rewritten as:


F_(E)=8.99(10)^(9) Nm^(2)/C^(2)(6(10)^(-8) C^(2))/((1 m)^(2)) (4)

Then, the new electrostatic force will be:


F_(E)= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

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