Question:

Two charges are placed on the x-axis. Charge Q1 = +2 μC is located at x = -3 m, and charge Q2 = -4 μC is located at x = +5 m.

a) Determine the electric field at a point on the x-axis located at x = +2 m due to these two charges.

b) Determine the direction of the electric field at that point.

c) If a positive test charge of 1 μC is placed at this point, what force will it experience?

Given:

• Electric constant, ε₀ = 8.854 x 10⁻¹² C²/N·m²

Answer:

a) To determine the electric field at a point on the x-axis located at x = +2 m, we need to calculate the electric field due to each charge individually using the formula:

Electric field, E = (k * Q) / r²

Where, k = Electric constant, ε₀ = 8.854 x 10⁻¹² C²/N·m² Q = Charge r = Distance between the charge and the point where electric field is calculated.

Calculating the electric field due to Q1: Q₁ = +2 μC = +2 x 10⁻⁶ C r₁ = distance = x - x₁ = 2 m - (-3 m) = 2 m + 3 m = 5 m

Substituting the values in the formula, E₁ = (k * Q₁) / r₁² E₁ = (8.854 x 10⁻¹² C²/N·m² * +2 x 10⁻⁶ C) / (5 m)²

Calculating the electric field due to Q2: Q₂ = -4 μC = -4 x 10⁻⁶ C r₂ = distance = x - x₂ = 2 m - (+5 m) = 2 m - 5 m = -3 m

Substituting the values in the formula, E₂ = (k * Q₂) / r₂² E₂ = (8.854 x 10⁻¹² C²/N·m² * -4 x 10⁻⁶ C) / (-3 m)²

The total electric field at point x = 2 m, E_total, is given by the vector sum of E₁ and E₂:

E_total = E₁ + E₂

b) To determine the direction of the electric field at the point x = 2 m, we consider the signs of the individual electric fields:

• If both electric fields have the same direction, the total electric field will point in that direction.
• If the electric fields have opposite directions, the total electric field will point towards the side with the stronger field.

c) To determine the force experienced by a positive test charge of 1 μC placed at the point x = 2 m, we can use the formula:

Force (F) = Charge (q) x Electric Field (E)

Given: q = +1 x 10⁻⁶ C

Substituting the values in the formula, F = (1 x 10⁻⁶ C) x E_total

Note: The direction of the force will depend on the sign of the test charge and the direction of the electric field.

Please be aware that due to limitations in text format, the mathematical expressions are simplified and rounded off in this explanation. In actual calculations, it is important to use the exact values and proper units.