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

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg