Locations$\mathbf{A}\mathbf{=}\mathbf{<}\mathbf{a}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathbf{}\mathbf{and}\mathbf{}\mathbf{B}\mathbf{=}\mathbf{<}\mathbf{b}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}$are on the +x axis, as shown in Figure 16.61. Four possible expressions for the electric field along the x axis are given below. For each expression for the electric field, select the correct expression (1–8) for the potential difference${\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}$. In each case K is a numerical constant with appropriate units.

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\overrightarrow{\mathbf{E}}\mathbf{=}<\frac{K}{x^2},0,0> \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\overrightarrow{\mathbf{E}}\mathbf{=}<\frac{K}{x^3},0,0> \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\overrightarrow{\mathbf{E}}\mathbf{=}<\frac{K}{x},0,0> \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\overrightarrow{\mathbf{E}}\mathbf{=}<\mathrm{Kx},0,0> \\ \mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\mathbf{0} \\ \mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\mathbf{K}\left(a-b\right) \\ \mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\mathbf{K}\left(\frac{1}{a}-\frac{1}{b}\right) \\ \mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\mathbf{K}\left(\frac{1}{a^3}a-\frac{1}{b^3}b\right) \\ \mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{K}\left(b^2-a^2\right) \\ \mathbf{\left(}\mathbf{6}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\mathbf{K}\mathbf{}\mathbf{In}\left(\frac{b}{a}\right) \\ \mathbf{\left(}\mathbf{7}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\mathbf{K}\left(a^3-b^3\right) \\ \mathbf{\left(}\mathbf{8}\mathbf{\right)}\mathbf{}{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{K}\left(\frac{1}{a^2}-\frac{1}{b^2}\right) \end{gathered}$$

The given data can be listed below,

• The coordinates of point A are,
• The coordinates of point B are,

The electric field is the inverse of potential, or, to put it another way, electric field equals potential per unit length.

For this case the electric field is given as,

The potential difference between point A and point B is given by,

$∆\mathrm{V}=-\int \overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{I}}$

Integration over b to a. the above equation will become,

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{x}}$

To find the potential difference substitute values of electric field.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{x}} \\ =-\mathrm{K}{\int }_{\mathrm{b}}^{\mathrm{a}}\frac{1}{{\mathrm{x}}^2}\mathrm{dx} \\ =\mathrm{K}{\left[\frac{1}{\mathrm{x}}\right]}_{\mathrm{b}}^{\mathrm{a}} \\ =\mathrm{K}\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right] \end{gathered}$$

Thus, for case (a), the correct potential difference is expression (3)$\mathrm{K}\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right]$.

The electric field for case (b) is given by,

The potential difference between point A and point B is given by,

$∆\mathrm{V}=-\int \overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{I}}$

Integration over b to a. the above equation will become,

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{x}}$

Substitute the value of electric field in the above equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{x}} \\ =-\mathrm{K}{\int }_{\mathrm{b}}^{\mathrm{a}}\frac{1}{{\mathrm{x}}^2}\mathrm{dx} \\ =\frac{1}{2}\mathrm{K}{\left[\frac{1}{{\mathrm{x}}^2}\right]}_{\mathrm{b}}^{\mathrm{a}} \\ =\frac{1}{2}\mathrm{K}\left[\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\right] \end{gathered}$$

Thus, for case (b), the correct potential difference is expression (8)$\frac{1}{2}\mathrm{K}\left[\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\right]$.

The electric field for case (c) is given by,

The potential difference between point A and point B is given by,

$∆\mathrm{V}=-\int \overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{I}}$

Integration over b to a. the above equation will become,

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{x}}$

Substitute the value of electric field in the above equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\frac{K}{x}dx \\ =-\mathrm{K}{\left[\mathrm{In}\mathrm{x}\right]}_{\mathrm{b}}^{\mathrm{a}} \\ =KIn\left(\frac{b}{a}\right) \end{gathered}$$

Thus, for case (c), the correct potential difference is expression (6)$KIn\left(\frac{b}{a}\right)$.

The electric field for case (c) is given by,

The potential difference between point A and point B is given by,

$∆\mathrm{V}=-\int \overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{I}}$

Integration over b to a. the above equation will become,

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{x}}$

Substitute the value of electric field in the above equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\overrightarrow{E}.d\overrightarrow{x} \\ =-{\int }_{\mathrm{b}}^{\mathrm{a}}Kxdx \\ =-K{\left[\frac{x^2}{2}\right]}_b^a \\ =\frac{1}{2}K\left(b^2-a^2\right) \end{gathered}$$

Thus, for case (d), the correct potential difference is expression (5) $\frac{1}{2}K\left(b^2-a^2\right)$.