Electric Force and Field Calculation: Coulomb's Law Problem

Published on June 12, 2026

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Solve for electric force and field intensity using Coulomb's Law, involving charges, distance, and electric fields.

Question

Resuelve el siguiente problema en un documento de word

Answer

Based on the image provided, here is the step-by-step solution to the physics problem.

Problem Data

  • Charge A ( ):
  • Charge B ( ):
  • Initial distance ( ):
  • Coulomb’s Constant ( ):

Part 1

1.1 Calculate the electric force between charges and .

We use Coulomb’s Law:
Substitute the values:
Answer: The electric force is .

1.2 Indicate if these ions attract or repel and explain why.

Answer: The ions attract each other.
Explanation: According to the laws of electrostatics, charges with opposite signs (one positive and one negative) exert an attractive force on each other.
Answer: Ion A moves to the right.
Explanation: Since the force is attractive, ion A is pulled toward ion B. Looking at the diagram, ion B is to the right of ion A, so ion A moves to the right.

Part 2

New Scenario:
  • Distance from A to P ( ):
  • Distance from B to P ( ):

2.1 Determine the intensity of the electric field at point P.

The electric field produced by a charge at distance is:
1. Field due to Ion A ( ):
Since is positive, the field at point P (to its right) points to the right.
2. Field due to Ion B ( ):
Since is negative, the field at point P (to its left) points to the right (towards the negative charge).
3. Total Electric Field ( ):
Since both vectors point to the right, we add their magnitudes:

2.2 Based on the result, what is the direction of the electric field at point P: left or right? Why?

Answer: The direction is to the right.
Explanation: Both the field from the positive charge (which pushes away from A) and the field from the negative charge (which pulls towards B) point in the same direction (to the right) at point P. Therefore, their vectors sum up in that direction.