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).

Question

Quizplus · Accepted Answer

To find the electric field, E, we need to take the negative gradient of the potential, V: E = -∇V Taking the partial derivative of V with respect to x, y, and z, we get: ∂V/∂x = 3²y² - 2z³ ∂V/∂y = x² + z³ ∂V/∂z = y³ - 2x³ At the point (x,y,z) = (1.0, 1.0, 1.0), these partial derivatives become: ∂V/∂x = 3²1² - 2³ ∂V/∂y = 1² + 1³ ∂V/∂z = 1³ - 2³1³ Plugging these values into the formula for the electric field, we get: E = -∇V = -(∂V/∂x)i - (∂V/∂y)j - (∂V/∂z)k E = -(3²1² - 2³)i - (1² + 1³)j - (1³ - 2³1³)k E = -5i + 2j + 5k The magnitude of this electric field vector is: |E| = sqrt((-5)^2 + 2^2 + 5^2) = sqrt(54) = 2sqrt(6) Rounding to one decimal place, we get: |E| ≈ 2.0 V/m Therefore, the best choice is D.

In a Certain Region, the Electric Potential Due to a Charge