Question

Two 2.00-µC point charges are located on the x axis. One is at x = 1.00 m, and the other is at x = -1.00 m. (a) Determine the electric field on the y axis at y = 0.500 m. (b) Calculate the electric force on a -3.00-µC charge placed on the y axis at y = 0.500 m

Answer 1

A) In the given point each charge creates the field:

$$E=\frac{1}{4\pi\epsilon_0}\frac{q}{x^2+y^2}.$$

The resulting field according to the superposition principle is

$$E_0=2E\cdot \text{cos}(\text{tan}^{-1}(x/y))=16099.689\text{ N/C}.$$

B) The electric force will be

$$F=QE_0=-0.048 \text{ N}.$$

The sign "-" means that the -3 microcoulomb charge will attract to the point in the middle between the 2 microcoulomb charges.