29: Problem 29.44

INTRO: The electric potential in a region of space is V=350V⋅m /√(x²+y²), where x and y are in meters.

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

SOLUTION: 
350⋅(x²+y²)^-1/2
The formula for this is E=|∇V|
∇V=<∂V/∂x, ∂V/∂y>
∂V/∂x = -350x(x²+y²)^-3/2
∂V/∂y = -350y(x²+y²)^-3/2
E(x,y)=<-350x(x²+y²)^-3/2,-350y(x²+y²)^-3/2>
E(3,3) = <-175/9√2,-175/9√2>
|E| = 175/9 V/m = 19 V/m

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

SOLUTION:
E(3,3) = <175/9√2,175/9√2>
since both sides are the same, it forms a perfect right triangle resulting in θ=45°