PART ONE:

A charge of 4.70μC is at the origin and a second charge of −2.85μC is on the positive x-axis 1.08 m from the origin.
Find the magnitude of the electric field at a point P on the y-axis 0.640 m from the origin. The value of the Coulomb constant is 8.99×10^9 N·m^2/C^2.
Answer in units of N/ C

Part Two:
Determine the direction of this electric field(as an angle between
−180◦ and 180◦ measured from the positive x-axis, with counter- clockwise positive).Answer in units of ◦

Part One: To find the magnitude of the electric field at point P on the y-axis, we can use the principle of superposition. The electric field created by each charge individually can be calculated using Coulomb's law, and then we can sum up their contributions.

The electric field due to a point charge can be found using the formula:

E = k * (Q / r^2)

where E is the electric field, k is the Coulomb constant (8.99×10^9 N·m^2/C^2), Q is the charge, and r is the distance between the charge and the point where the field is being measured.

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

For the charge at the origin (Q1 = 4.70μC), the distance to point P (r1) is 0.640 m. Plugging these values into the equation, we get:

E1 = (8.99×10^9 N·m^2/C^2) * (4.70×10^-6 C) / (0.640 m)^2

Calculate the value for E1.

For the charge on the positive x-axis (Q2 = -2.85μC), the distance to point P (r2) is 1.08 m. Plugging these values into the equation, we get:

E2 = (8.99×10^9 N·m^2/C^2) * (-2.85×10^-6 C) / (1.08 m)^2

Calculate the value for E2.

The total electric field at point P is the vector sum of the individual electric fields. Since both charges are on the x-axis, their electric fields will point along the x-axis. Therefore, the y-components of the electric fields due to each charge will cancel each other out.

Thus, the magnitude of the electric field at point P is equal to the sum of the absolute values of the x-components of the electric fields:

|E| = |E1x| + |E2x|

Calculate the value for |E|.

Part Two: To determine the direction of the electric field at point P, we can use trigonometry. Since the electric fields due to each charge only have x-components and point along the x-axis, their contribution to the total electric field will also lie along the x-axis.

Therefore, the direction of the resultant electric field can be described by the angle it makes with the positive x-axis. We can use the arctangent function to find this angle. The formula is:

θ = arctan(Ey/Ex)

In this case, since both Ey and Ex are zero, the angle will also be zero degrees.

Hence, the direction of the electric field at point P is 0° degrees, measured from the positive x-axis.