29: Problem 29.43

INTRO: The electric potential in a region of space is V=(180 x²− 280 y²) V, where x and y are in meters.

PART A:
What is the strength of the electric field at (x,y)=(1.0m,3.0m) ?
Express your answer using two significant figures.

SOLUTION: 
The formula for this is E=∇V
∇V=<∂V/∂x, ∂V/∂y>
∂V/∂x = 360x
∂V/∂y = -560y
E(x,y)=<360x,-560y>
E(1,3) = <360,-1680>
to get the magnitude of the vector, √(360)²+(1680)²|
E = 1718 V/m = 1700 V/m

PART B: What is the direction of the electric field at (x,y)=(1.0m,3.0m) ? 
Give the direction as an angle (in degrees) counterclockwise from the positive x-axis.

SOLUTION:
so, with the vector that we found before, E(1,3) = <360,-1680>, we can initially determine the angle that it is pointing from below the positive x axis
the 'adjacent' side of the triangle this vector creates is 360
the 'opposite' side is 1680
tan(1680/360) would solve the angle in that right triangle
θ=22°
now the question asks for the direction as an angle counterclockwise from the positive x axis, so the exact opposite way that we just determined. Since the electric field points the opposite way [E=-dV/dx], our angle isn't in the 4th quadrant it's actually in the 2nd. 
E = <-360, 1680>
θ=102°