Particle 1 carrying -4.0 μC of charge is fixed at the origin of an xy coordinate system, particle 2 carrying +8.0 μC of charge is located on the x axis at x = 4.0 m , and particle 3, identical to particle 2, is located on the x axis at x = -4.0 m .
To proceed with the question, we need to find the electric field at two points: point P on the positive x-axis, and point Q on the negative x-axis.
To find the electric field at point P, we need to consider the electric field contributions from particle 2 and particle 3 separately. The electric field E1 at point P due to particle 2 can be calculated using Coulomb's law:
E1 = k * |q1| / r^2
where k is the electrostatic constant (k ≈ 9 × 10^9 N*m^2/C^2), |q1| is the magnitude of the charge on particle 1 (8.0 μC = 8.0 × 10^-6 C), and r is the distance between particle 2 and point P (4.0 m). Since particle 2 carries a positive charge, the electric field points outward from the charge.
Similarly, the electric field E2 at point P due to particle 3 can be calculated in the same manner. However, since particle 3 is negatively charged, the direction of the electric field will be towards the charge.
To find the total electric field at point P, we need to add the contributions from both particles:
E_total = E1 + E2
Now, let's determine the electric field at point Q on the negative x-axis. The process will be the same as for point P, but in this case, the electric field due to particle 2 will be towards the charge, and the electric field due to particle 3 will be outward from the charge:
E_total_Q = E1_Q + E2_Q
Finally, remember that the electric field follows the principle of superposition, meaning that the total electric field at a point due to multiple charges is the vector sum of the individual electric fields at that point.
With this information, you can proceed to calculate the electric fields at points P and Q.