Electric field and potential: In a certain region, the electric potential due to a charge distribution is given by the equation V(x,y,z) = 3x²^y² + yz³ - 2z³^x, 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).

Question

Quizplus · Accepted Answer

To find the electric field vector, E, we need to take the negative gradient of the potential function, V. 
That is, E = -∇V.
Taking partial derivatives of V with respect to x, y, and z:
∂V/∂x = 3S1U1P12S1S1P0S1U1P1yS1S1P0S1U1P12S1S1P0 - 2S1U1P13S1S1P0S1U1P1xS1S1P0
∂V/∂y = 3S1U1P12S1S1P0S1U1P1xS1S1P0
∂V/∂z = yS1U1P13S1S1P0 - 2S1U1P13S1S1P0
Thus, at position (1.0, 1.0, 1.0), we have:
∂V/∂x = 3(1)S1(1)U1P12S1S1P0S1U1P11(1)S1S1P0S1U1P12S1S1P0 - 2S1(1)U1P13S1S1P0S1U1P11(1)S1S1P0 = 3S1U1P12S1S1P0S1U1P1S1S1P0S1U1P12S1S1P0 - 2S1U1P13S1S1P0S1U1P1S1S1P0 = S1(3-2)U1P12S1S1P0S1U1P1S1S1P0S1U1P12S1S1P0 = S1U1P13S1S1P0S1U1P12S1S1P0
∂V/∂y = 3S1U1P12S1S1P0S1U1P11(1)S1S1P0 = 3S1U1P12S1S1P0S1U1P1S1S1P0
∂V/∂z = (1)(1)S1U1P13S1S1P0 - 2S1U1P13S1S1P0 = -1S1U1P13S1S1P0
So the electric field vector at (1.0, 1.0, 1.0) is:
E = -(∂V/∂x)i - (∂V/∂y)j - (∂V/∂z)k
= -S1U1P13S1S1P0S1U1P12S1S1P0 i - 3S1U1P12S1S1P0S1U1P1S1S1P0 j + S1U1P13S1S1P0 k
Now we need to plug in the specific values at (1.0, 1.0, 1.0):
E = -S1U1P13S1S1P0S1U1P12S1S1P0 i - 3S1U1P12S1S1P0S1U1P1S1S1P0 j + S1U1P13S1S1P0 k
= -S1(1)(1)S1(1)U1P13S1S1P0S1U1P12S1S1P0 i - 3S1(1)U1P12S1S1P0S1U1P1(1)S1S1P0 j + S1(1)U1P13S1S1P0 k
= -S1/2 i - 3S1/2 j - S1/2 k
The magnitude of the electric field vector is:
|E| = sqrt((S1/2)^2 + (3S1/2)^2 + (S1/2)^2)
= sqrt(11/2)
~= 3.72 V/m
Among the given answer choices, D) 8.6 V/m is the closest to our result. Therefore, the correct answer is D).

Electric Field and Potential: in a Certain Region, the Electric