An electric potential is given by V=4x² + 2y² + x + z + 2 volts.What is the electric field at the point (2,2,2)?

Question

Quizplus · Accepted Answer

The electric field $\vec{E}$ is related to the electric potential $V$ by the negative gradient of $V$, $\vec{E} = -\nabla V$. Given $V = 4x^2 + 2y^2 + x + z + 2$, the components of the electric field are found by taking the partial derivatives of $V$ with respect to $x$, $y$, and $z$, and negating them: $E_x = -\frac{\partial V}{\partial x} = -8x - 1$, $E_y = -\frac{\partial V}{\partial y} = -4y$, and $E_z = -\frac{\partial V}{\partial z} = -1$. Substituting $x = 2$, $y = 2$, and $z = 2$ gives $E_x = -17$, $E_y = -8$, and $E_z = -1$, so the electric field at the point (2,2,2) is $-17\hat{i} - 8\hat{j} + \hat{k}$ N/C.

An Electric Potential Is Given by V=4x2 + 2y2

An Electric Potential Is Given by V=4x² + 2y²