Answer
$E = 150~N/C$
Work Step by Step
We can find $E_x$:
$E_x = -\frac{dV}{dx}$
$E_x = -2.00yz^2$
$E_x = (-2.00)(-2.00)(4.00)^2$
$E_x = 64.0~N/C$
We can find $E_y$:
$E_y = -\frac{dV}{dy}$
$E_y = -2.00xz^2$
$E_y = (-2.00)(3.00)(4.00)^2$
$E_y = -96.0~N/C$
We can find $E_z$:
$E_z = -\frac{dV}{dz}$
$E_z = -4.00xyz$
$E_z = (-4.00)(3.00)(-2.00)(4.00)$
$E_z = 96.0~N/C$
We can find the magnitude of the electric field:
$E = \sqrt{()^2+()^2+()^2}$
$E = \sqrt{(64.0~N/C)^2+(-96.0~N/C)^2+(96.0~N/C)^2}$
$E = 150~N/C$
