Question

In a certain region, the electric potential due to a charge distribution is given by the equation v(x,y,z) = 3x2y2 + yz3 - 2z3x, where x, y, and z are measured in meters and v is in volts. calculate the magnitude of the electric field vector at the position (x,y,z) = (1.0, 1.0, 1.0).

Answer 1

Magnitude = 11.05

Explanation:

The electric potential due to a charge distribution is given by

(Equation 1)

V(x,y,z)=3x²y² + yz³ - 2z³x.......(1)

Differentiating the above equation 1 with respect to x,

(Equation 2)

Vx= 6xy² + 2z³

Differentiating the above equation 1 with respect to y,

(Equation 3)

Vy= 6x²y + z³

Differentiating the above equation 1 with respect to z,

(Equation 4)

Vz= 3yz² - 6z²x

Substituting the value of x,y and z in equations 2,3 and 4

Vx= 6+2=8

Vy=6+1=7

Vz=3-6= -3

Magnitude = √((Vx)²+(Vy)²+(Vz)²)

Magnitude = √(64)+(49)+(9)

Magnitude=√122

Magnitude = 11.05

Answer 2

The electric potential due to a charge distribution is given by

`V(x,y,z)=3x^(2)y^(2)+yz^(3)+2z^(3)x` (Equation 1)

Differentiating the above equation 1 with respect to x,

`V_(x)=6xy^(2)+2z^(3)` (Equation 2)

Differentiating the above equation 1 with respect to y,

`V_(y)=6x^(2)y+z^(3)` (Equation 3)

Differentiating the above equation 1 with respect to z,

`V_(z)=3yz^(2)+6z^(2)x` (Equation 4)

Substituting the value of x,y and z in equations 2,3 and 4

`image`

`V_(y)=6+1=7`

`V_(z)=3+6=9`

Magnitude =
`\sqrt{8^(2)+7^(2)+9^(2)}`

=
`√(194)`

= 14.0