Question

Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran

encima de una recta, Determine: a) la distancia que se deben separar las
cargas q1 y q2 para que la fuerza eléctrica entre ambas cargas sea de 10
N , b) a que distancia la fuerza se duplica c) a que distancia la fuerza se
quintuplica

Answer 1

To determine the separation distance between two charges for a given electric force, we can use Coulomb's Law. Substituting the given values into the equation, we find that the distance is approximately 0.424 m.

Step-by-step explanation:

To determine the distance that the charges q1 and q2 need to be separated for the electric force between them to be 10 N, we can use Coulomb's Law. The equation for Coulomb's Law is:

F = k * (|q1 * q2| / r^2)

where F is the electrostatic force, q1 and q2 are the magnitudes of the charges, r is the distance between the charges, and k is the electrostatic constant (k ≈ 8.99 * 10^9 Nm^2/C^2).

By substituting the given values into the equation, we can solve for r. The distance is:

r = √((k * |q1 * q2|) / F) ≈ √((8.99 * 10^9 Nm^2/C^2 * |(-50 * 10^-6 C) * (30 * 10^-6 C)|) / 10 N)

r ≈ 0.424 m

Answer 2

Datos:

q1 = -50 μC =
`50*10^(-6)`

q2 = +30 μC =
`30*10^(-6)`

F = 10 N

a) x si la F = 10N

Aplicando la Ley de Coulomb:

x =
`\sqrt(k0 * q1 *q2)/(F)` =
`\sqrt((9*10^(9) )*(50*10^(-6))*(30*10^(-6)))/(10)` = 1,162m

b) x si la F = 20 N

x=
`\sqrt((9*10^(9) )*(50*10^(-6))*(30*10^(-6)))/(20)` = 0,822m

c)x si la F = 50 N

x =
`\sqrt((9*10^(9) )*(50*10^(-6))*(30*10^(-6)))/(50)` = 0,520m