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3 years ago

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1. Una carga Q1 = + 12 μC se coloca a una distancia r = 0.024 m desde una carga Q2 = + 16 μC. a) Determina la magnitud de la fue

rza electrostática que actúa sobre las dos cargas, Q1 y Q2. b) ¿Es la fuerza la atracción o repulsión? 2. Determina la intensidad del campo eléctrico a una distancia radial de r = 48 mm desde una carga de Q = 24 μC. 3. Una carga Q1 = 24 mC se coloca a una distancia r = 0.032 m desde una carga Q2 = - 12 μC. a. Determina la cantidad de energía potencial eléctrica que tiene la carga Q1. b. Determina el potencial eléctrico en la posición de Q2.

1 answer:

lyudmila [28]3 years ago

3 0

Answer:

1. a. 3,000 N

b. Repulsión

2. 46.875 × 10⁶ N/C

3. a. 81,000 J

b. 6.75 × 10⁹ V

Explanation:

1. Los parámetros dados son;

Q₁ = +12 μC, Q₂ = +16 μC

La distancia entre las cargas, r = 0.024

La magnitud de la fuerza electrostática, F, entre cargas se da como sigue;

$F = k \times \dfrac{Q_1 \cdot Q_2}{r^2}$

Donde, k = constante de Coulomb = 9.0 × 10⁹ N · m² / C²

Por lo tanto, obtenemos;

F = 9.0 × 10⁹ × 12 × 10⁻⁶ × 16 × 10⁻⁶ / 0.024² = 3.000

La magnitud de la fuerza electrostática, entre las cargas, F = 3000 N

(b) Dado que tanto Q₁ como Q₂ son cargas positivas, y las cargas iguales se repelen entre sí, la fuerza es la repulsión.

2) La intensidad de un campo eléctrico, E, se da como sigue;

$E = \dfrac{k \cdot Q}{r^2}$

La magnitud de la carga, Q = 24 μC

La distancia donde se mide el campo, r = 48 mm = 0.048 m

Por lo tanto, E = 9.0 × 10⁹ × 12 × 10⁻⁶ / 0.048² = 46,875,000 N / C

La intensidad de un campo eléctrico, E = 46,875,000 N / C = 46.875 × 10⁶ N / C

3. La magnitud de las cargas son;

Q₁ = 24 mC

Q₂ = -12 μC

La distancia entre las cargas, r = 0.032 m

un. El potencial eléctrico de una carga, $U_E$ , se da de la siguiente manera;

$U_E = k \times \dfrac{Q_1 \cdot Q_2}{r}$

Por lo tanto;

$U_E$ = 9.0×10⁹ × 24 × 10⁻³ × (-12) × 10⁻⁶ /0.032 = -81,000

La energía potencial eléctrica entre la carga, Q₁ y Q₂= -81,000 J

b. El potencial eléctrico de Q₁ en Q₂, V₁ = $k \times \dfrac{Q_1 }{r}$

Por lo tanto, V₁ = 9.0×10⁹ × 24 × 10⁻³/0.032 = 6.75 × 10⁹

El potencial eléctrico de Q₁ en Q₂, V₁ = 6.75 × 10⁹ V

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