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
2 answers:
Answer:
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
Answer:
Like the other person said, the answer is 2.8 x 10^9
Explanation:
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