The electric potential of a charge distribution is given by the equation V(x) = 3 x² y² + y z³ - 2 z³x, where x, y, z are measured in meters and V is measured in volts. Calculate the magnitude of the electric field vector at the position (x,y,z) = (1.0, 1.0, 1.0).

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Quizplus · Accepted Answer

The electric field vector is given by the negative gradient of the electric potential:$$\overrightarrow{E} = -
abla V = -\left(\frac{\partial V}{\partial x}\hat{x} + \frac{\partial V}{\partial y}\hat{y} + \frac{\partial V}{\partial z}\hat{z}ight)$$Plugging in the given electric potential, we get:$$\overrightarrow{E} = -6xy^2\hat{x} - 6x^2y\hat{y} - (3y^2z^2 + 6xz^2)\hat{z}$$At the position (x,y,z) = (1.0, 1.0, 1.0), the electric field vector becomes:$$\overrightarrow{E} = -6(1.0)(1.0)^2\hat{x} - 6(1.0)^2(1.0)\hat{y} - (3(1.0)^2(1.0)^2 + 6(1.0)(1.0)^2)\hat{z}$$$$\overrightarrow{E} = -6\hat{x} - 6\hat{y} - 9\hat{z}$$The magnitude of the electric field vector is:$$|\overrightarrow{E}| = \sqrt{(-6)^2 + (-6)^2 + (-9)^2} = \sqrt{81} = 9 V/m$$Therefore, the correct answer is D) 8.6 V/m.

The Electric Potential of a Charge Distribution Is Given by the Equation