Two point charges, q1 and q2 are separated by a distance d. For convenience, we will say that q1 is at the origin (0,0) on a coordinate system, q2 is at (d,0) and a reference point p is directly above q2 at a distance h at point (d,h). q1 has value 1μ C and q2 has value 8μ C. Take d= 1.1m and h= 0.7m. Note k=9×109Nm2/C2 What is the direction of the electric field at point p, as measured from a line parallel to the horizontal axis? Give your answer in degrees and to 2 significant digits.

To solve this problem, we first need to calculate the electric field at point P due to each charge separately, and then combine these to find the total electric field.

1. Calculate the electric field due to q1 at P (E1):

The magnitude of the electric field E1 is given by Coulomb's law: E = k*q/r^2, where k is Coulomb's constant, q is the charge and r is the distance from the charge.

In this case, q1 = 1μC = 1*10^-6 C, r1 = sqrt(d^2 + h^2) = sqrt((1.1m)^2 + (0.7m)^2) = 1.3m.

So, E1 = kq1/r1^2 = 910^9 Nm^2/C^2 * 1*10^-6 C / (1.3m)^2 = 5.3 N/C.

The direction of E1 is from q1 to P, which forms an angle θ1 with the horizontal axis. We can find θ1 using trigonometry: tan(θ1) = h/d = 0.7m/1.1m, so θ1 = arctan(0.7/1.1) = 32.47 degrees.

2. Calculate the electric field due to q2 at P (E2):

Similarly, we find that q2 = 8μC = 8*10^-6 C, r2 = h = 0.7m.

So, E2 = kq2/r2^2 = 910^9 Nm^2/C^2 * 8*10^-6 C / (0.7m)^2 = 145.71 N/C.

The direction of E2 is from q2 to P, which is vertically upwards, or 90 degrees from the horizontal axis.

3. Combine E1 and E2 to find the total electric field at P:

The total electric field E at P is the vector sum of E1 and E2. Since E1 and E2 are not in the same direction, we need to add them as vectors.

The horizontal component of E is E1*cos(θ1) = 5.3 N/C * cos(32.47 degrees) = 4.48 N/C.

The vertical component of E is E1*sin(θ1) + E2 = 5.3 N/C * sin(32.47 degrees) + 145.71 N/C = 3.5 N/C + 145.71 N/C = 149.21 N/C.

So, the magnitude of E is sqrt(E_horizontal^2 + E_vertical^2) = sqrt((4.48 N/C)^2 + (149.21 N/C)^2) = 149.3 N/C.

The direction of E forms an angle θ with the horizontal axis. We can find θ using trigonometry: tan(θ) = E_vertical/E_horizontal = 149.21 N/C / 4.48 N/C, so θ = arctan(149.21/4.48) = 88.22 degrees.

So, the direction of the electric field at point P, as measured from a line parallel to the horizontal axis, is 88.22 degrees.