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Kalyan ramji sain, of india, had a mustache that measured 3.39 m from end to end in 1993. suppose two charges, q and 3q, are placed 3.39 m apart. if the magnitude of the electric force between the charges is 2.4 × 10−6 n, what is the value of q?

Respuesta :

mavila18

Answer:

[tex]q=3.19*10^{-8}C[/tex]

Explanation:

The expression for the electric force between two charges

[tex]F=k\frac{q_1q_2}{r^2}\\[/tex]

where K is the Coulomb's constant (k=9*10^{9}Nm^2/C^2), and we have that

q1=q

q2=3q

r=3.39m

F=2.4*10^{-6}N

By replacing in the formula we have

[tex]F=2.4*10^{-6}N=(9*10^{9}\frac{Nm^2}{C^2})\frac{(q)(3q)}{(3.39m)^2}\\\\[/tex]

and by taking apart q we have

[tex]q=3.19*10^{-8}C[/tex]

hope this helps!!

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