Two point charges each experience a 1-n electrostatic force when they are 2 cm apart. if they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them? a. 2 nb. 1/8 nc. 1/16 nd. 1/4 ne. 1/2 n

The right option is C. 1/16 N

Explanation:

From coulomb's law,

F = kqq'/r² Equation 1

Where F = force between the charges, q = first charge, q' = second charge, r = distance of separation between the the charges, k = constant of proportionality.

qq' = Fr²/k Equation 2

Given: F = 1 N, r = 2 cm = 0.02 m

Substituting into equation 2, to get the value of the product of the charges in terms of k

qq' = 1×0.02²/k

qq' = 0.0004/k Equation 3

When they are moved to a new separation of 8 cm,

Then r = 8 cm = 0.08 m

F = kqq'/0.08² Equation 4

Substituting the value of qq' in equation 3 into equation 4

F = k(0.0004k)/0.0064

F = 4/64

F = 1/6 N

Hence the electric force on each of the is 1/16 N

The right option is C. 1/16 N

answera compound;

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