Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

Question

A positive charge of magnitude Q₁ = 0.45 nC is located at the origin. A negative charge Q₂ = -9.5 nC is located on the positive x-axis at x = 6.5 cm from the origin. The point P is located y = 7.5 cm above charge Q₂. I have no clue where to start.

Answer 1

The question asks what the x component of the electric field vector at point P is due to charge Q1. According to this, you neglect the contribution of charge Q2.

Assume there is a small positive charge located at point P. By definition the magnitude of the electric field at point P due to charge Q1 is

E = kQ1/d² where k is the coulomb constant and d is the straight line distance from Q1 to P.

The distance is the hypotenuse of the triangle formed by Q1, Q2, and P.

d = √0.065² + 0.075² = 0.099 m at an angle of θ = arctan 7.5/6.5 = 49°.

|E| = 8.99 x 10⁹ Nm² C^-2 x 0.45 x 10-⁹ C/ 0.099² m² = 412 n/C

The x component would be Ex = 412 x cos 49° = 270 N/C

Please check my math.