What is the magnitude of the electric field at the point (8.70 - 9.10 7.20 ) m if the electric potential is given by V

This question is incomplete, the complete question is;

What is the magnitude of the electric field at the point (8.70i - 9.10j + 7.20k ) m if the electric potential is given by V = 4.30xyz², where V is in volts and x, y  and z are in meters.

the magnitude of the electric field is 5648.67 N/C

Explanation:

Given the data in the question;

Suppose electric field at this point is;

E = Exi + Eyj + Ezk

Now, electric field is given by;

E = -dV/dr

so,

Ex = -d( 4.30xyz² )/dx = -4.30yz²

Ey = -d( 4.30xyz² )/dy = -4.30xz²

Ez = -d( 4.30xyz² )/dz = -8.60xyz

so

E = -4.30yz² i  - 4.30xz² j - 8.60xyz k

now, at the point (8.70i - 9.10j + 7.20k )  

E = (-4.30(-9.10)(7.20)²) i  + (-4.30(8.70)(7.20)²) j + (-8.60(8.70)(-9.10)(7.20) k

E =  2028.499 i - 1939.334 j + 4902.206 k

so, Magnitude of electric field will be;

|E| = √( Ex² +  Ey² + Ez² )

we substitute

|E| = √( (2028.499)² +  (-1939.334)² + (4902.206)² )

|E| = √( 31907448.222993 )

|E| = 5648.67 N/C

Therefore, the magnitude of the electric field is 5648.67 N/C

the earths atmosphere is made up of nitrogen and oxygen. solids can turn into gasses through a process called sublimation. sublimation is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.

hope this : d