Question

9. Over a certain region of space, the electric potential is V = . (a) Find the expressions for the x, y, and z components of the electric field over this region. (b) What is the magnitude of the field at the point P that has coordinates (1.00, 0, 22.00) m?

Answer 1

Answer:

a)

The expression for the electric potential in this problem is:

$V=5x-3x^2y+2yz$

where

x, y, z are the three spatial coordinates

The relationship between components of the electric field and electric potential is:

$E_x=-\frac{dV}{dx}\\E_y=-\frac{dV}{dy}\\E_z=-\frac{dV}{dz}$

Therefore, we have to calculate the derivatives of the potential over the three variables.

Doing so, we find:

$E_x=-\frac{d}{dx}(5x-3x^2y+2yz)=-(5-6xy)=6xy-5$

$E_y=-\frac{d}{dy}(5x-3x^2y+2yz)=-(-3x^2+2z)=3x^2-2z$

$E_z=-\frac{d}{dz}(5x-3x^2y+2yz)=-(2y)=-2y$

b)

Here we want to find the magnitude of the electric field at the point P that has coordinates

P (1.00, 0, 22.00) m

First of all, we find the components of the electric field at that point by substituting

x = 1.00

y = 0

z = 22.0

We find:

$E_x=6xy-5=6(1)(0)-5=-5 N/C\\E_y=3x^2-2z=3(1)^2-2(22)=-41 N/C\\E_z=-2y=-2(0)=0$

Now, the magnitude of the electric field is given by

$E=\sqrt{E_x^2+E_y^2+E_z^2}$

And by substituting,

$E=\sqrt{(-5)^2+(-41)^2+0}=41.3 N/C$