Question

A dipole is oriented along the x axis. The dipole moment is $\mathit{p}\mathbf{=}\mathbf{\left(}\mathit{q}\mathit{s}\mathbf{\right)}$.

(a) Calculate exactly the potential $\mathit{V}$(relative to infinity) at a location $\left\{x,0,0\right\}$on the $\mathit{x}$axis and at a location $\left\{0,y,0\right\}$on the $\mathit{y}$axis, by superposition of the individual $\mathbf{1}\mathbf{/}\mathit{r}$contributions to the potential.

(b) What are the approximate values of $\mathit{V}$at the locations in part (a) if these locations are far from the dipole?

(c) Using the approximate results of part (b), calculate the gradient of the potential along the $\mathit{x}$axis, and show that the negative gradient is equal to the x component ${\mathit{E}}_{\mathbf{x}}$ of the electric field.

(d) Along the y axis, $\mathit{d}\mathit{V}\mathbf{/}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$. Why isn’t this equal to the magnitude of the electric field $\mathit{E}$along the $\mathit{y}$axis?

Short answer

It is showed that the gradient along $y$axis is not equal to the magnitude of the electric field $E$along the$y$ axis.

Step-by-step solution

Write the given data from the question.

The dipole is oriented along the $x$axis.

The dipole, $p=qs$

Along the y axis, $\mathit{d}\mathit{V}\mathbf{/}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$.

Determine the formulas to calculate the potential and gradient of the potential.

The expression to calculate the potential at the location is given as follows.

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{\overrightarrow{r}}$

Here,$\overrightarrow{r}$ is the relative position vector.

Along the y axis, dV/dy=0. Why isn’t this equal to the magnitude of the electric field Ealong the y axis.

Since the potential along the $y$axis is zero. Therefore, the gradient of the potential along the $y$is also zero and the gradient along axis is not equal to the magnitude of the electric field $E$along the $y$axis.

Hence it is showed that the gradient along $y$axis is not equal to the magnitude of the electric field $E$along the$y$ axis.