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Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be

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Answer:

F' = F/4

Thus, the magnitude of electrostatic force will become one-fourth.

Explanation:

The magnitude of force applied by each charge on one another can be given by Coulomb's Law:

F = kq₁q₂/r²   -------------- equation 1

where,

F = Force applied by charges

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between the charges

Now, in the final state the charges on both spheres are halved. Therefore,

q₁' = q₁/2

q₂' = q₂/2

Hence, the new force will be:

F' = kq₁'q₂'/r²

F' = k(q₁/2)(q₂/2)/r²

F' = (kq₁q₂/r²)(1/4)

using equation 1:

F' = F/4

Thus, the magnitude of electrostatic force will become one-fourth.

The magnitude of the electrostatic force will be F' = F/4

The magnitude of the electrostatic force:

Here we used Coulomb's Law:

F = kq₁q₂/r²   -------------- equation 1

Here

F = Force applied by charges

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between the charges

Now

q₁' = q₁/2

q₂' = q₂/2

So, the new force should be

F' = kq₁'q₂'/r²

F' = k(q₁/2)(q₂/2)/r²

F' = (kq₁q₂/r²)(1/4)

So,

F' = F/4

Learn more about force here: https://brainly.com/question/14282312

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