Answer:
a)
The expression for the electric potential in this problem is:
![V=5x-3x^2y+2yz](https://img.qammunity.org/2021/formulas/physics/high-school/km5gghcz7lta59u61ecu0n8l04dj9tsfy7.png)
where
x, y, z are the three spatial coordinates
The relationship between components of the electric field and electric potential is:
![E_x=-(dV)/(dx)\\E_y=-(dV)/(dy)\\E_z=-(dV)/(dz)](https://img.qammunity.org/2021/formulas/physics/high-school/q37pbpbji6aprhfru00eytb5y0a5xq52ge.png)
Therefore, we have to calculate the derivatives of the potential over the three variables.
Doing so, we find:
![E_x=-(d)/(dx)(5x-3x^2y+2yz)=-(5-6xy)=6xy-5](https://img.qammunity.org/2021/formulas/physics/high-school/b18er0x7fjuexfz9tt6r99ahv7kit0i4kk.png)
![E_y=-(d)/(dy)(5x-3x^2y+2yz)=-(-3x^2+2z)=3x^2-2z](https://img.qammunity.org/2021/formulas/physics/high-school/xw7bqseczvda3xas7kjjvu41uwm2t8cgu7.png)
![E_z=-(d)/(dz)(5x-3x^2y+2yz)=-(2y)=-2y](https://img.qammunity.org/2021/formulas/physics/high-school/ivod453pjd6wh6sn5td284vuwr2ciooi2r.png)
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](https://img.qammunity.org/2021/formulas/physics/high-school/sdk9yexv10xibo7glpg9pywwjnm5phegb8.png)
Now, the magnitude of the electric field is given by
![E=√(E_x^2+E_y^2+E_z^2)](https://img.qammunity.org/2021/formulas/physics/high-school/uieby33olhek9inl9pdoo3cqpjkvi2mc64.png)
And by substituting,
![E=√((-5)^2+(-41)^2+0)=41.3 N/C](https://img.qammunity.org/2021/formulas/physics/high-school/6hugha6i9i6mre13t0549sfkgrpac0ojxp.png)