Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

3. Dos cargas puntuales [tex]q_1=3 \times 10^6 \, C[/tex] y [tex]q_2=4 \times 10^{-5} \, C[/tex] están separadas 0.5 m y ubicadas en el vacío. Calcule el valor de la fuerza entre las cargas.

Sagot :

Para calcular la fuerza entre dos cargas puntuales en el vacío, utilizamos la ley de Coulomb, que se expresa como:

[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]

Donde:
- [tex]\( F \)[/tex] es la magnitud de la fuerza entre las dos cargas.
- [tex]\( k \)[/tex] es la constante de Coulomb, [tex]\( k \approx 8.9875 \times 10^9 \, N \cdot m^2 \cdot C^{-2} \)[/tex].
- [tex]\( q_1 \)[/tex] y [tex]\( q_2 \)[/tex] son las magnitudes de las dos cargas puntuales.
- [tex]\( r \)[/tex] es la distancia entre las dos cargas.

Dado:
- [tex]\( q_1 = 3 \times 10^6 \, C \)[/tex]
- [tex]\( q_2 = 4 \times 10^{-5} \, C \)[/tex]
- [tex]\( r = 0.5 \, m \)[/tex]

Sigamos los pasos para calcular la fuerza:

1. Plug in the given values into Coulomb's law.

[tex]\[ q_1 = 3 \times 10^6 \, C \][/tex]
[tex]\[ q_2 = 4 \times 10^{-5} \, C \][/tex]
[tex]\[ r = 0.5 \, m \][/tex]
[tex]\[ k \approx 8.9875 \times 10^9 \, N \cdot m^2 \cdot C^{-2} \][/tex]

2. Substitute these values into the formula:

[tex]\[ F = 8.9875 \times 10^9 \frac{|3 \times 10^6 \cdot 4 \times 10^{-5}|}{(0.5)^2} \][/tex]

3. Calculate the product of the charges:

[tex]\[ |3 \times 10^6 \cdot 4 \times 10^{-5}| = 3 \times 10^6 \cdot 4 \times 10^{-5} \][/tex]
[tex]\[ = 12 \times 10^1 \, C^2 \][/tex]
[tex]\[ = 1.2 \times 10^2 \, C^2 \][/tex]

4. Square the distance:

[tex]\[ (0.5)^2 = 0.25 \, m^2 \][/tex]

5. Calculate the force:

[tex]\[ F = 8.9875 \times 10^9 \frac{1.2 \times 10^2}{0.25} \, N \][/tex]

6. Simplify the fraction:

[tex]\[ \frac{1.2 \times 10^2}{0.25} = 4.8 \times 10^2 \, \text{N} \][/tex]

7. Finally, multiply by [tex]\( k \)[/tex]:

[tex]\[ F = 8.9875 \times 10^9 \times 4.8 \times 10^2 \, N \][/tex]

This gives:

[tex]\[ F \approx 4314024857936.7246 \, N \][/tex]

Por lo tanto, la magnitud de la fuerza entre las cargas es aproximadamente [tex]\( 4314024857936.7246 \)[/tex] Newtons (N).
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.