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igor_vitrenko [27]

2 years ago

12

- Calcular la magnitud de la fuerza eléctrica entre dos cargas cuyos valores son: q1 = 3.5 coulomb, q2 = 6 coulomb, al estar sep

aradas en el vacío por una distancia de 0,4 m.

1 answer:

mario62 [17]2 years ago

8 0

Answer:

FE = 1.17*10^12 N

Explanation:

The magnitude of the electric force is given by the following formula:

$F_E=k\frac{q_1q_2}{r^2}$ ( 1 )

q1 = 3.5C

q2 = 6C

r = 0.4m

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

You replace the values of q1, q2, r and k in the expression (1):

$F_E=(8.98*10^9Nm^2/C^2)\frac{(3.5C)(6C)}{(0.4m)^2}\\\\F_E=1.17*10^{12}N$

hence, the magnitude of the electric force is 1.17*10^12 N

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A uniformly charged rod (length = 2.0 m, charge per unit length = 5.0 nc/m) is bent to form one quadrant of a circle. what is th

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The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

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