A +5.00 nC charge is placed on the x-axis at x = 1.00 m, and a -6.00 nC point charge is placed on the y-axis at y = -2.00 m. What is the direction of the electric field at the origin as measured counterclockwise from the positive x-axis?

The electric field E created by a point charge q at a distance r from the charge is given by the formula:

E = k*q/r^2

where k is Coulomb's constant (8.9910^9 Nm^2/C^2).

The direction of the electric field is always from positive to negative charges.

Let's calculate the electric field at the origin due to each charge:

1. For the +5.00 nC charge at x = 1.00 m:

E1 = kq1/r1^2 = (8.9910^9 Nm^2/C^2) * (5.0010^-9 C) / (1.00 m)^2 = 44.95 N/C

The direction of this electric field is along the positive x-axis.

2. For the -6.00 nC charge at y = -2.00 m:

E2 = kq2/r2^2 = (8.9910^9 Nm^2/C^2) * (-6.0010^-9 C) / (2.00 m)^2 = -67.43 N/C

The direction of this electric field is along the positive y-axis because the electric field direction is from positive to negative charges.

The total electric field at the origin is the vector sum of these two electric fields.

Since the electric fields are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the total electric field:

E_total = sqrt(E1^2 + E2^2) = sqrt((44.95 N/C)^2 + (-67.43 N/C)^2) = 80.38 N/C

The direction of the total electric field is the arctangent of the ratio of the two fields:

θ = arctan(E2/E1) = arctan((-67.43 N/C) / (44.95 N/C)) = -56.31 degrees

Since the angle is measured counterclockwise from the positive x-axis, the direction of the total electric field at the origin is 56.31 degrees below the positive x-axis.