Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them

F' = F/12

Therefore, the electrostatic force is reduced to one-twelve of its original value.

Explanation:

The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:

F = kq₁q₂/r²   equation 1

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between charges

Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:

F' = kq₁'q₂'/r'²

where,

q₁' = (1/3)q₁

q₂' = q₂

r' = 2r

Therefore,

F' = k(1/3 q₁)(q₂)/(2r)²

F' = (1/12)kq₁q₂/r²

using equation 1:

F' = F/12

Therefore, the electrostatic force is reduced to one-twelve of its original value.

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.