Chapter 26, Problem 55PQ

The electric potential is given by V = 4x²z + 2xy² − 8yz² in a region of space, with x, y, and z in meters and V in volts.

1. a. What are the x, y, and z components of the electric field in this region?
2. b. What is the magnitude of the electric field at the coordinates (2.00 m, −2.00 m, 1.00 m)?

To determine

The x, y and z components of electric field in the region.

Answer to Problem 55PQ

The electric field in the regionis $\overrightarrow{E}=-\left(8xz+2y^2\right)\widehat{i}-\left(4xy-8z^2\right)\widehat{j}-\left(4x^2-16yz\right)\widehat{k}$.

Explanation of Solution

Write the equation for the electric field.

    $\overrightarrow{E}=E_x\widehat{i}+E_y\widehat{j}+E_z\widehat{k}$                                                                                          (I)

Here, $E_x$ is the x component of the electric field, $E_y$ is the y component of the electric field and $E_z$ is the z component of the electric field.

Write the equation for the the x component of the electric field.

    $E_x=-\frac{\partial V}{\partial x}$                                                                                                         (II)

Here, $E_x$ is the electric field along x axis and $\partial V/\partial x$ is the partial derivative of the potential gradient along x axis.

Write the equation for the the y component of the electric field.

    $E_y=-\frac{\partial V}{\partial y}$                                                                                                        (III)

Here, $E_y$ is the electric field along y axis and $\partial V/\partial y$ is the partial derivative of the potential gradient along y axis.

Write the equation for the the z component of the electric field.

    $E_z=-\frac{\partial V}{\partial z}$                                                                                                        (IV)

Here, $E_z$ is the electric field along z axis and $\partial V/\partial z$ is the partial derivative of the potential gradient along z axis.

Conclusion:

Substitute $4x^{2}z+2x{y}^2-8y{z}^2$ for $V$ in equation (II) to find $E_x$.

    $\begin{array}{c}{E}_x=-\frac{\partial \left(4x^{2}z+2x{y}^2-8y{z}^2\right)}{\partial x}\\ =-\left(8xz+2y^2\right)\end{array}$

Substitute $4x^{2}z+2x{y}^2-8y{z}^2$ for $V$ in equation (III) to find $E_y$.

    $\begin{array}{c}{E}_y=-\frac{\partial \left(4x^{2}z+2x{y}^2-8y{z}^2\right)}{\partial y}\\ =-\left(4xzy-8y{z}^2\right)\end{array}$

Substitute $4x^{2}z+2x{y}^2-8y{z}^2$ for $V$ in equation (IV) to find $E_z$.

    $\begin{array}{c}{E}_z=-\frac{\partial \left(4x^{2}z+2x{y}^2-8y{z}^2\right)}{\partial z}\\ =-\left(4x^2-16yz\right)\end{array}$

Thus, from equation (I), write the electric field.

    $\overrightarrow{E}=-\left(8xz+2y^2\right)\widehat{i}-\left(4xy-8z^2\right)\widehat{j}-\left(4x^2-16yz\right)\widehat{k}$

Thus, the electric field in the region is $\overrightarrow{E}=-\left(8xz+2y^2\right)\widehat{i}-\left(4xy-8z^2\right)\widehat{j}-\left(4x^2-16yz\right)\widehat{k}$.

To determine

The magnitude of the electric field.

Answer to Problem 55PQ

The magnitude of the electric field is $58.8\text{V/m}$.

Explanation of Solution

Write the equation for the the x component of the electric field.

    $E_x=-\left(8xz+2y^2\right)$                                                                                          (V)

Write the equation for the the y component of the electric field.

    $E_y=-\left(4xzy-8y{z}^2\right)$                                                                                     (VI)

Write the equation for the the z component of the electric field.

    $E_z=-\left(4x^2-16yz\right)$                                                                                       (VII)

Write the equation for the magnitude of the electric field.

    $E=\sqrt{E_x^2+E_y^2+E_z^2}$                                                                                      (VIII)

Here, $E$ is the total magnitude of the electric field.

Conclusion:

Substitute $2.00\text{m}$ for $x$, $-2.00\text{m}$ for $y$ and $1.00\text{m}$ for $z$ in equation (V) to find $E_x$.

    $\begin{array}{c}{E}_x=-\left(8\left(2.00\text{m}\right)\left(1.00\text{m}\right)+2{\left(-2.00\text{m}\right)}^2\right)\\ =-24.0\text{V/m}\end{array}$

Substitute $2.00\text{m}$ for $x$, $-2.00\text{m}$ for $y$ and $1.00\text{m}$ for $z$ in equation (VI) to find $E_y$.

    $\begin{array}{c}{E}_y=-\left(4\left(2.00\text{m}\right)\left(1.00\text{m}\right)\left(-2.00\text{m}\right)-8\left(-2.00\text{m}\right){\left(1.00\text{m}\right)}^2\right)\\ =24.0\text{V/m}\end{array}$

Substitute $2.00\text{m}$ for $x$, $-2.00\text{m}$ for $y$ and $1.00\text{m}$ for $z$ in equation (VII) to find $E_z$.

    $\begin{array}{c}{E}_z=-\left(4{\left(2.00\text{m}\right)}^2-16\left(-2.00\text{m}\right)\left(1.00\text{m}\right)\right)\\ =-48.0\text{V/m}\end{array}$

Substitute $-24.0\text{V/m}$ for $E_x$, $24.0\text{V/m}$ for $E_y$ and $-48.0\text{V/m}$ for $E_z$ in equation (VII) to find $E$.

    $\begin{array}{c}E=\sqrt{{\left(-24.0\text{V/m}\right)}^2+{\left(24.0\text{V/m}\right)}^2+{\left(-48.0\text{V/m}\right)}^2}\\ =58.8\text{V/m}\end{array}$

Thus, the magnitude of the electric field is $58.8\text{V/m}$.