91Ó°ÊÓ

Deriving V from E requires integration, and deriving E from V requires differentiation.

So by using the equation 23.19 for each component of E and drive equation (1), get the electric field for the three components where equation 23.19 gives the partial derivative of the potential to obtain the electric field.

For $E_x$:

$$\begin{gathered} E_x=-\frac{∂V}{∂x} \\ {E}_x=-\frac{∂V}{∂x}(Axy-B{x}^2+Cy) \\ {E}_x=-Ay+2Bx \end{gathered}$$

Forlocalid="1664278262524" $E_y$:

$$\begin{gathered} E_y=-\frac{∂V}{∂y} \\ {E}_y=-\frac{∂V}{∂y}(Axy-B{x}^2+Cy) \\ {E}_y=-Ax-C \end{gathered}$$

For $E_z$:

$$\begin{gathered} E_z=-\frac{∂V}{∂z} \\ {E}_z=-\frac{∂}{∂z}(Axy-B{x}^2+Cy) \\ {E}_z=0 \end{gathered}$$

Using the equation (2), (3) and (4) to get zero electric field E, where

$E_x=E_y=E_z=0$

For z components, as shown by equation (4) the electric field is zero at every point on z axis. For y components, as shown by equation (3), the electric field will be zero in y component when $x=\left(\frac{-C}{A}\right)$

.For x components, as shown by equation (2), the electric field is zero at $x=\left(\frac{-C}{A}\right)$but in this case y value will be

$$\begin{gathered} E_x=-Ay+2Bx \\ 0=-Ay+2Bx \\ y=\frac{2Bx}{A} \\ y=\left(\frac{-2Bx}{A^2}\right) \end{gathered}$$

In general, the electric field is zero at $x=\left(\frac{-C}{A}\right)y=\left(\frac{-2Bx}{A^2}\right)$ and any point on z.