What would be the electrostatic force between two charges of equal magnitude if the distance between them is halved

Biology

2 answers:

dalvyx [7]3 years ago

4 0

Answer:

the forces between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them

Natalka [10]3 years ago

4 0

Answer:

The force becomes 4 times as much when the distance is halved.

Explanation:

F1 = k*q1 * q2/r^2 = kq^2/r^2

F2 = kq^2/1 / (r/2)^2

Set up a proportion using F1/F2 as you reference.

Notice that you don't know any number that can be used in this equation. However k q^2 does not change as you move the two charges closer together. That being so, you can still show what happens to F1 and F2.

F1/F2 = 1/r^2 // 1/(r^2/4)

Take the denominator, invert the fraction and multiply. That creates another cancellation.

F1/F2 = 1/r^2 * r^2 /4 1

F1/F2 = r^2 // r^2/4 which allows you to cancel r^2

F1/F2 = 1/4 Now you need to interpret F1 and F2. Cross Multiply

4F1 = F2

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