A dipole is centered at the origin, with its axis along the y axis, so that at locations on the y axis, the electric field due to the dipole is given by

$\overrightarrow{\mathbf{E}}\mathbf{=}⟨\mathbf{0}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{2}\mathbf{qs}}{{\mathbf{y}}^{\mathbf{3}}}\mathbf{,}\mathbf{0}⟩\frac{\mathbf{\text{V}}}{\mathbf{\text{m}}}$

The charges making up the dipole are$\mathbf{+}\mathbf{3}\mathbf{\text{\hspace{0.17em}nC}}$ and $\mathbf{-}\mathbf{3}\mathbf{\text{\hspace{0.17em}nC}}$, and the dipole separation is$\mathbf{2}\mathbf{\text{\hspace{0.17em}mm}}$ (Figure 16.82). What is the potential difference along a path starting at location$\mathbf{⟨}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{.03}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{\text{\hspace{0.17em}m}}$ and ending at location$\mathbf{⟨}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{.04}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{\text{\hspace{0.17em}m}}$ ?

The given data is listed below,

• First charge making up the dipole is,${\mathrm{q}}_1=3\text{\hspace{0.17em}nC}=3\text{\hspace{0.17em}nC}\times \frac{1\text{\hspace{0.17em}C}}{1\times {10}^9\text{nC}}=3\times {10}^{-9}\text{\hspace{0.17em}C}$
• Second charge making up the dipole is,${\mathrm{q}}_2=-3\text{\hspace{0.17em}nC}=-3\text{\hspace{0.17em}nC}\times \frac{1\text{\hspace{0.17em}C}}{1\times {10}^9\text{nC}}=-3\times {10}^{-9}\text{\hspace{0.17em}C}$
• The initial point of path is,$\mathrm{A}=⟨0,0.03,0⟩\text{\hspace{0.17em}m}$
• The end point of the path is,$\mathrm{B}=⟨0,0.04,0⟩\text{\hspace{0.17em}m}$
• The separation between dipole is, $\mathrm{s}=2\text{\hspace{0.17em}mm}=2\text{\hspace{0.17em}mm}\times \frac{1\text{\hspace{0.17em}m}}{1000\text{\hspace{0.17em}mm}}=2\times {10}^{-3}\text{\hspace{0.17em}m}$.
• The electric field is,$\overrightarrow{E}=⟨0,\frac{2Kqs}{y^3},0⟩\text{\hspace{0.17em}V/m}$

It is a metric that measures the strength of the dipole. It is a vector quantity that travels along the axis from negative to positive charge.

The product of the magnitudes of charge and dipole separation determines the magnitude of dipole moment.

The potential difference due to electric field in $y$-direction is given by,

$\mathrm{\Delta V}=-\underset{{\mathrm{q}}_1}{\overset{{\mathrm{q}}_2}{\int }}\mathrm{Edy}$

Here,$E$is the given electric field,${\mathrm{q}}_1$is the first charge making up the dipole, and$q_2$is the second charge making up the dipole.

Substitute the value of electric field in the above expression.

$\mathrm{\Delta V}=-{\int }_{0.03\text{\hspace{0.17em}m}}^{0.04\text{\hspace{0.17em}m}}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2\mathrm{qs}}{{\mathrm{y}}^3}\right)\mathrm{dy}$

Here,$\frac{1}{4\pi {\epsilon }_0}$is the electric constant with value$\mathrm{k}=(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)$.

Substitute all the values in the above expression.

$\begin{array}{c}\Delta V=-{\int }_{0.03\text{\hspace{0.17em}m}}^{0.04\text{\hspace{0.17em}m}}k\frac{2qs}{y^3}dy\\ =-2kqs{\int }_{0.03\text{\hspace{0.17em}m}}^{0.04\text{\hspace{0.17em}m}}\frac{1}{y^3}dy\\ =-2kqs{\left[-\frac{1}{2y^2}\right]}_{0.03}^{0.04}\\ =2(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(3\times {10}^{-9}\text{\hspace{0.17em}C})(2\times {10}^{-3}\text{\hspace{0.17em}m}){\left[\frac{1}{2y^2}\right]}_{0.03\text{\hspace{0.17em}m}}^{0.04\text{\hspace{0.17em}m}}\\ =2(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(3\times {10}^{-9}\text{\hspace{0.17em}C})(2\times {10}^{-3}\text{\hspace{0.17em}m})\left(\frac{1}{2{\left(0.04\text{\hspace{0.17em}m}\right)}^2}-\frac{1}{2{\left(0.03\text{\hspace{0.17em}m}\right)}^2}\right)\\ =-26.24\text{\hspace{0.17em}N}\cdot \text{m}/\text{C}\times \frac{1\text{\hspace{0.17em}V}}{1\text{\hspace{0.17em}N}\cdot \text{m}/\text{C}}\\ =-26.24\text{\hspace{0.17em}V}\end{array}$

Thus, the value of potential difference of dipole is$-26.24\text{\hspace{0.17em}V}$ .