A point charge of +5.00 nC is placed on the x axis at x = 1.00 m and another point charge of -6.00 nC is placed on the y axis at y = -2.00 m. Determine the direction of the electric field at the origin in degrees counterclockwise about the positive x-axis.
To determine the direction of the electric field at the origin, we need to calculate the contributions of the electric fields created by the two point charges at the origin and then find the resultant direction.
The electric field due to a point charge is given by the equation:
E = k * (q / r^2)
where E is the electric field magnitude, k is the electrostatic constant (8.99 × 10^9 N m²/C²), q is the charge magnitude, and r is the distance between the charge and the point where the electric field is being calculated.
Let's start by calculating the electric field due to the +5.00 nC charge at the origin (0, 0). Since this charge is placed on the x-axis at x = 1.00 m, the distance between the charge and the origin is r1 = 1.00 m.
Using the equation for electric field, the magnitude of the electric field due to the +5.00 nC charge is:
E1 = (8.99 × 10^9 N m²/C²) * (5.00 × 10^(-9) C) / (1.00 m)²
E1 = 44.95 N/C
Now let's calculate the electric field due to the -6.00 nC charge at the origin (0, 0). Since this charge is placed on the y-axis at y = -2.00 m, the distance between the charge and the origin is r2 = 2.00 m.
Using the equation for electric field, the magnitude of the electric field due to the -6.00 nC charge is:
E2 = (8.99 × 10^9 N m²/C²) * (-6.00 × 10^(-9) C) / (2.00 m)²
E2 = -26.965 N/C
Since the electric field is a vector quantity, we need to take into account the direction of the electric fields created by the two charges. The electric field due to the +5.00 nC charge points radially outward from the charge, and the electric field due to the -6.00 nC charge points radially inward toward the charge.
To find the resultant electric field at the origin, we need to add the electric fields due to the two charges. Since the electric fields have different magnitudes and opposite directions, we can subtract their magnitudes to find the resultant electric field's magnitude:
E_total = |E1| - |E2|
E_total = |44.95 N/C| - |26.965 N/C|
E_total = 17.985 N/C
The direction of the electric field at the origin can be determined by finding the angle counterclockwise from the positive x-axis. We can use trigonometry to calculate this angle:
θ = atan(E2 / E1)
θ = atan(-26.965 N/C / 44.95 N/C)
θ = atan(-0.599)
θ ≈ -30.08 degrees
However, since we want the angle counterclockwise about the positive x-axis, we need to convert this negative angle to a positive angle:
θ = 360 degrees + (-30.08 degrees)
θ ≈ 329.92 degrees
Therefore, the direction of the electric field at the origin is approximately 329.92 degrees counterclockwise about the positive x-axis.