Question

Coulomb’s Law describes the force between two (and only two!) charged objects. Suppose Object 1 has a charge of +2.71 μC and Object 2 has a charge of +18.5 μC. Object 2 is located a distance of 0.04480 m to the right of Object 1 (Yes — please break down and draw a quick sketch!). What is the magnitude and direction of the electrical force on Object 2 due to Object 1?

What is the magnitude and direction of the electrical force on Object 1 due to Object 2?

You now move Object 2 such that it is located a distance of 0.08960 m to the right of Object 1 (twice the distance as before). What is the magnitude and direction of the electrical force on Object 2 due to Object 1?

The new force on Object 1 due to Object 2 (at a separation of 0.08960 m) is ____ times the original force on Object 1 due to Object 2 (at a separation of 0.04480 m)? (Provide a numerical answer to fill in the blank.)

Answer 1

a )

using

F = k q₁ q₂ / d²

= 9 x 10⁹ x 2.71 x 10^-6 x 18.5 x 10^-6 / 0.0448²

force on object 2 due to 1 is F = 224.8161 N to right

b )

now

force on object 1 due to 2 = 9 x 10⁹ x 2.71 x 10^-6 x 18.5 x 10^-6 / 0.0448²

force on object 1 due to 2 is F = 224.8161 N to left

c )

for twice distance is

force on object 2 due to 1 F = 9 x 10⁹ x 2.71 x 10^-6 x 18.5 x 10^-6 / 0.0896²   

= 56.204 N

d )

the distance is doubled the force is (1/2)²

= 1/4 times

so the new force is 0.25 times than the previous force