What is the electric force acting between two charges of −0.0085 C and −0.0025 C that are 0.0020 m apart?

Physics

2 answers:

dsp733 years ago

3 0

$F=k \frac{ Q_{1} Q_{2} }{ r^{2} }$
where k=8.99e9
So (8.99e9)(-0.0085)(-0.0025)/(0.002)^2=4.7759e10N

katen-ka-za [31]3 years ago

3 0

Answer:

4.8•10^10 N

Explanation:

(9.00•10^9)(-0.0085)(-0.0025)

————————————————= 4.8•10^10

0.0020^2

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