Question

The electric potential in a region of space is V=( 260 x2? 150 y2)V, where x and y are in meters.

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.

I keep getting 331 degress. I have tried 30 and 210 as well, but they are all wrong..

Answer 1

The electric field intensity vector is derived by taking the negative gradient of the potential field.
E = -?V = -(?V/?x a_x + ?V/?y a_y + ?V/?z a_z)
E = -(520x a_x - 300y a_y)
E(2,3) = - (520*3a_x - 300*3a_y) V/m
E(2,3) = -1560a_x + 900a_y V/m

a. ||E(2,3)|| = sqrt(1560^2 + 900^2)
||E(2,3)|| = 1800.1 V/m

b. ? = tan-1(Ey/Ex)
? = tan-1(900/(-1560))
? = -30 deg
? = 180 - 30
? = 150 deg (since x is negative and y is positive)