Three charges are on the y axis. Two of the charges, each --q, are located y = ±d, and the third charge, +2q, is located at y = 0. Derive an expression for the electric field at a point P on the x-axis.

Question

in simple steps

Answer 1

To derive an expression for the electric field at point P on the x-axis, we can use the principle of superposition. This principle states that the total electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge.

1. Divide the problem into two cases: one where the point P is to the left of the origin and one where the point P is to the right of the origin. This will help simplify the calculations.

Case 1: Point P is to the left of the origin
- In this case, the charges at y = ±d will create electric fields that are directed towards the positive x-axis and will have a magnitude of:
E1 = k*(q)/((d^2 + x^2)^1.5), where x is the distance from the charges at y = ±d to point P, and k is the Coulomb constant.

- The charge at y = 0 will create an electric field directed towards the negative x-axis and will have a magnitude of:
E2 = k*(2q)/((x^2)^1.5).

- To find the net electric field at point P, subtract the magnitude of E2 from the magnitude of E1: E = E1 - E2.

Case 2: Point P is to the right of the origin
- In this case, the charges at y = ±d will create electric fields that are directed towards the negative x-axis and will have a magnitude of:
E1 = k*(q)/((d^2 + x^2)^1.5), where x is the distance from the charges at y = ±d to point P, and k is the Coulomb constant.

- The charge at y = 0 will create an electric field directed towards the positive x-axis and will have a magnitude of:
E2 = k*(2q)/((x^2)^1.5).

- To find the net electric field at point P, add the magnitude of E2 to the magnitude of E1: E = E1 + E2.

2. Once you have the expression for the net electric field at point P, you can simplify it further by substituting the values of q, d, and k if given.

Note: The direction of the electric field at point P will depend on the signs of the charges. If q is positive, the field direction will be reversed.