Question

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.

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

Short answer

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

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

c) For case (c), the correct potential difference is expression (6) $\mathrm{K}\mathrm{In}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$.

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

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of the electric field.

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

(a) Determination of potential difference for case (a).

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)$\mathrm{K}\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right]$.

(b) Determination of potential difference for case (b)

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)$\frac{1}{2}\mathrm{K}\left[\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\right]$.

(c) Determination of potential difference for case (c)

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)$KIn\left(\frac{b}{a}\right)$.

(d) Determination of potential difference for case (d)

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) $\frac{1}{2}K\left(b^2-a^2\right)$.