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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).

Respuesta :

The electric potential due to a charge distribution is given by

[tex] V(x,y,z)=3x^{2}y^{2}+yz^{3}+2z^{3}x [/tex] (Equation 1)

Differentiating the above equation 1 with respect to x,

[tex] V_{x}=6xy^{2}+2z^{3} [/tex] (Equation 2)

Differentiating the above equation 1 with respect to y,

[tex] V_{y}=6x^{2}y+z^{3} [/tex] (Equation 3)

Differentiating the above equation 1 with respect to z,

[tex] V_{z}=3yz^{2}+6z^{2}x [/tex] (Equation 4)

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

[tex]
V_{x}=6+2=8 [/tex]

[tex] V_{y}=6+1=7 [/tex]

[tex] V_{z}=3+6=9 [/tex]

Magnitude = [tex] \sqrt{8^{2}+7^{2}+9^{2}} [/tex]

=[tex] \sqrt{194} [/tex]

= 14.0

Answer:

Magnitude = 11.05

Step-by-step 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

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