What is the electric force acting between two charges of -0.0045 c and -0.0025 c that are 0.0060 m apart? use fe=kq1q2/r^2 and k = 9.00 x 10^9 n*m^2/c^2 a. 1.7 x 10^7 n b. -1.7 x 10^7 n c. -2.8 x 10^9 n d. 2.8 x 10^9 n

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N