91Ó°ÊÓ

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,

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

The potential difference can be defined as follows,

$△V=\stackrel{→}{-E}△\stackrel{→}{I}$

The dot product of the Equation (2) is,

$△V=-\left|\stackrel{→}{E}\right|\left|\stackrel{→}{I}\right|\cos \mathrm{θ}$

Where, $\mathrm{θ}$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,

$â–³{\mathrm{U}}_{eI}=qâ–³V$

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

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

Therefore, The displacement vector $△\stackrel{→}{I}$ 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 ${90}^{∘}$. Substitute the values in Equation (3) as follows,

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

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.

$â–³{\mathrm{U}}_{eI}=qâ–³V$

$\begin{array}{l}â–³U_{eI}=q\left(0\right)\\ â–³U_{ei}=0\end{array}$

(4)

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

$â–³{\mathrm{U}}_{eI}=qâ–³V$

$\begin{array}{l}â–³U_{eI}=q\left(0\right)\\ â–³U_{eI}=0\end{array}$

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