As shown in Figure 16.72, three large, thin, uniformly charged plates are arranged so that there are two adjacent regions of uniform electric field. The origin is at the center of the central plate. Location A is $\mathbf{<}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{}\mathit{m}$, and location B is$\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{}\mathit{m}$ . The electric field$\overrightarrow{{\mathbf{E}}_{\mathbf{1}}}$ has the value $<725,0,0>\mathit{V}\mathbf{/}\mathit{m}$, and $\overrightarrow{{\mathbf{E}}_{\mathbf{2}}}$is $<-425,0,0>\mathbf{}\mathit{V}\mathbf{/}\mathit{m}$.

(d) What is the minimum kinetic energy the electron must have at location A in order to ensure that it reaches location B?

The location of $A$,

The location of $B$,

The electrical field,

The electrical field,

The charge of the electron, $q=1.6\times {10}^{-19}\text{C}$.

The expression to calculate the change in the potential is given as follows.

$\Delta V=\left(E_x\Delta x+E_y\Delta y+E_z\Delta z\right)$

Here, $E_x$is the electrical field in$x-$ direction, $\Delta x$is the change in the distance$x-$direction,$E_y$ is the electrical field in$y-$ direction,$\Delta y$ is the change in the distance$y-$ direction,$E_z$ is the electrical field in $z-$direction, $\Delta z$and is the change in the distance $z-$direction.

The expression to change in the kinetic energy is given as follows.

$\Delta K=q\Delta V$ …… (i)

(d)

The electron should have the enough kinetic energy to traverse the potential difference between A and B. therefore the minimum kinetic energy of the electron should be$-328\times {10}^{-19}\text{J}$ ,So the electron can reach at the location B.

Hence the minimum kinetic energy of the electron is $-328\times {10}^{-19}\text{J}$.