The electric potential is given by V = 6x - 8 | vedclass

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(D) The electric field components are given by $E_x = -\frac{\partial V}{\partial x}$,$E_y = -\frac{\partial V}{\partial y}$,and $E_z = -\frac{\partia

Vedclass · Accepted Answer

$20$

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(D) The electric field components are given by $E_x = -\frac{\partial V}{\partial x}$,$E_y = -\frac{\partial V}{\partial y}$,and $E_z = -\frac{\partial V}{\partial z}$.Calculating the partial derivatives:$E_x = -\frac{\partial}{\partial x}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(6 - 8y^2) = 8y^2 - 6$$E_y = -\frac{\partial}{\partial y}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(-16xy - 8 + 6z) = 16xy + 8 - 6z$$E_z = -\frac{\partial}{\partial z}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(6y - 8z) = 8z - 6y$At the origin $(x=0, y=0, z=0)$:$E_x = 8(0)^2 - 6 = -6 \ V/m$$E_y = 16(0)(0) + 8 - 6(0) = 8 \ V/m$$E_z = 8(0) - 6(0) = 0 \ V/m$The magnitude of the electric field is $E = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-6)^2 + 8^2 + 0^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ N/C$.The electric force is $F = qE = 2 \ C 	imes 10 \ N/C = 20 \ N$.

The electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Find the electric force in $N$ acting on a point charge of $2 \ C$ placed at the origin.

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