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).

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

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.