Questions: Item II: Desarrolle los siguientes problemas: 1. Dada dos cargas de 2,00 µC, como se muestra en la Figura 1, y una carga de prueba positiva q=1,28 × 10^-18 C colocada en el origen. a) ¿Cuál es la fuerza neta ejercida por las dos cargas 2,00 µC sobre la carga de prueba q ? b) ¿Cuál es el campo eléctrico en el origen debido a las dos cargas de 2,00 µ ? c) ¿Cuál es el potencial eléctrico en el origen debido a las dos cargas de 2,00 µC ?

Solution Steps

• Given:

  • Two charges of \( 2.00 \, \mu C \) located at \( x = -0.800 \, m \) and \( x = 0.800 \, m \).
  • Test charge \( q = 1.28 \times 10^{-18} \, C \) placed at the origin.

• Formula:

  • Coulomb's Law: \( F = k \frac{|q_1 q_2|}{r^2} \)
  • \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)

• Calculation:

  • Distance from each charge to the origin: \( r = 0.800 \, m \)
  • Force due to each charge: \[ F = k \frac{(2.00 \times 10^{-6} \, C) (1.28 \times 10^{-18} \, C)}{(0.800 \, m)^2} \] \[ F = 8.99 \times 10^9 \frac{(2.00 \times 10^{-6}) (1.28 \times 10^{-18})}{0.64} \] \[ F = 8.99 \times 10^9 \times 4 \times 10^{-24} \] \[ F = 3.596 \times 10^{-14} \, N \]

• Net Force:

  • The forces due to each charge are equal in magnitude but opposite in direction along the x-axis.
  • Therefore, the net force on \( q \) is zero.

• Formula:

  • Electric field due to a point charge: \( E = k \frac{|q|}{r^2} \)

• Calculation:

  • Electric field due to each charge at the origin: \[ E = k \frac{2.00 \times 10^{-6} \, C}{(0.800 \, m)^2} \] \[ E = 8.99 \times 10^9 \frac{2.00 \times 10^{-6}}{0.64} \] \[ E = 8.99 \times 10^9 \times 3.125 \times 10^{-6} \] \[ E = 2.809375 \times 10^4 \, N/C \]

• Net Electric Field:

  • The electric fields due to each charge are equal in magnitude but opposite in direction along the x-axis.
  • Therefore, the net electric field at the origin is zero.

• Formula:

  • Electric potential due to a point charge: \( V = k \frac{q}{r} \)

• Calculation:

  • Electric potential due to each charge at the origin: \[ V = k \frac{2.00 \times 10^{-6} \, C}{0.800 \, m} \] \[ V = 8.99 \times 10^9 \frac{2.00 \times 10^{-6}}{0.800} \] \[ V = 8.99 \times 10^9 \times 2.5 \times 10^{-6} \] \[ V = 2.2475 \times 10^4 \, V \]

• Net Electric Potential:

  • The potentials due to each charge are scalar quantities and add up.
  • Therefore, the net electric potential at the origin is: \[ V_{\text{net}} = 2 \times 2.2475 \times 10^4 \, V = 4.495 \times 10^4 \, V \]

Final Answer