Locations A, B , and C are in a region of uniform electric field, as shown in Figure 16.65. Location A is at (-0.5,0,0)m. Location Bis at (-0.5,0,0)m . In the region the electric field has the value (750,0,0)N/C. (a)For a path starting at Band ending at C, calculate: (1) the displacement vector$△\overrightarrow{I}$ , (2) the change in electric potential, (3) the potential energy change for the system when a proton moves from B to C, (4) the potential energy change for the system when an electron moves from Bto C, (b) Which of the following statements are true in this situation? Choose all that are correct. (1) the potential difference cannot be Choose zero because the electric field is not zero along this path, (2) when a proton moves along this path, the electric force does zero network on the proton, (3) $△\overrightarrow{I}$ is perpendicular to $\overrightarrow{E}$.

Consider that in a uniform electric field the locations are A, B and C. Location A and B is at (-0.5,0,0) m and (-0.5,0,0) m respectively. The electric field has the value (750,0,0) N/C. Consider that the path exists between the location B and C .

The potential difference is the potential developed from the work done to move the particular charge from some point to the origin.

1.

Consider the path starting at B(-0.5,0,0) mand ending at C. Since the point C is perpendicular to the axis AB thus the point C has the same x component values and the y is unknown that is below point B. Thus, the location C is (0.5, -y, 0) m.

The displacement vector is defined as follows,

$$\begin{gathered} △\overrightarrow{l}=(△x,△y,△z) \\ △\overrightarrow{l}=\left(\left(x_f-x_i\right),\left(y_f-y_i\right),\left(z_f-z_i\right)\right) \end{gathered}$$

The potential difference can be defined as follows,

$△V=\overrightarrow{-E}△\overrightarrow{I}$

The dot product of the Equation (2) is,

$△V=-\left|\overrightarrow{E}\right|\left|\overrightarrow{I}\right|\cos \theta$

Where, $\theta$represents the angle between the direction of the charged particle and the electric field direction.

From the Equation (3), the change in potential energy can be found as follows,

$△{\mathrm{U}}_{eI}=q△V$

Using the Equation (1), the displacement vector $\overrightarrow{I}$can be found as follows,

$\begin{array}{l}\overrightarrow{I}=\{0.5,-y,0\}-\{0.5,0,0\}\\ \overrightarrow{I}=\{0,-y,0\}\mathrm{m}\end{array}$

Therefore, The displacement vector $△\overrightarrow{I}$ is $\left(0,-y,0\right)m$ .

(2)

From the Equation (2), the angle between the direction of the charged particle and the electric filed is ${90}^{\circ }$. Substitute the values in Equation (3) as follows,

$\begin{array}{l}△V=-\left|E\right|\left|I\right|\cos {90}^{\circ }\\ △V=-\left|E\right|\left|I\right|\left(0\right)\\ △V=0\end{array}$

Therefore, The change in electric potential is zero.

(3)

Since the potential difference between the points is zero, thus the potential energy for proton to from point B and C is zero.

$△{\mathrm{U}}_{eI}=q△V$

$\begin{array}{l}△U_{eI}=q\left(0\right)\\ △U_{ei}=0\end{array}$

(4)

Since the potential difference between the points is zero, thus the potential energy for electron to from point B and C is zero.

$△{\mathrm{U}}_{eI}=q△V$

$\begin{array}{l}△U_{eI}=q\left(0\right)\\ △U_{eI}=0\end{array}$

Therefore, The potential energy change for the system when a electron moves from B to C is zero.