how did you find Φdipole​=4*πϵ*0​1​*r*3**p**⋅**r**​

The expression for the electric potential \(\Phi_{\text{dipole}}\) due to a dipole is derived from the fundamental definition of the potential created by a dipole in electrostatics.

### Derivation of the Potential \(\Phi_{\text{dipole}}\)

An electric dipole consists of two equal and opposite charges \(+q\) and \(-q\) separated by a small distance \(d\). The dipole moment \(\mathbf{p}\) is defined as:

\[
\mathbf{p} = q\mathbf{d}
\]

where \(\mathbf{d}\) is the vector pointing from the negative charge to the positive charge.

### Potential Due to a Point Charge

The potential \(\Phi\) at a point \(\mathbf{r}\) due to a point charge \(q\) located at \(\mathbf{r}'\) is given by:

\[
\Phi(\mathbf{r}) = \frac{1}{4\pi \epsilon_0} \frac{q}{|\mathbf{r} - \mathbf{r}'|}
\]

### Potential Due to a Dipole

For a dipole, the potential at a point \(\mathbf{r}\) in space due to the positive charge at \(\mathbf{r}_+\) and the negative charge at \(\mathbf{r}_-\) is:

\[
\Phi(\mathbf{r}) = \frac{1}{4\pi \epsilon_0} \left( \frac{+q}{|\mathbf{r} - \mathbf{r}_+|} - \frac{-q}{|\mathbf{r} - \mathbf{r}_-|} \right)
\]

Assuming that the separation \(d\) between the charges is small compared to the distance \(|\mathbf{r}|\) to the observation point, we can approximate the distance as:

\[
|\mathbf{r} - \mathbf{r}_\pm| \approx r \mp \hat{\mathbf{r}} \cdot \mathbf{d}/2
\]

where \(\hat{\mathbf{r}} = \mathbf{r}/r\) is the unit vector in the direction of \(\mathbf{r}\).

Expanding the expression in a Taylor series and keeping only the first-order terms, the potential simplifies to:

\[
\Phi(\mathbf{r}) \approx \frac{q}{4\pi \epsilon_0} \left( \frac{1}{r - \hat{\mathbf{r}} \cdot \mathbf{d}/2} - \frac{1}{r + \hat{\mathbf{r}} \cdot \mathbf{d}/2} \right)
\]

\[
\Phi(\mathbf{r}) \approx \frac{q}{4\pi \epsilon_0} \left( \frac{\hat{\mathbf{r}} \cdot \mathbf{d}}{r^2} \right)
\]

Since \(\mathbf{p} = q\mathbf{d}\), this becomes:

\[
\Phi(\mathbf{r}) = \frac{1}{4\pi \epsilon_0} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^2}
\]

Finally, expressing the potential in terms of the position vector \(\mathbf{r}\), we get:

\[
\Phi(\mathbf{r}) = \frac{1}{4\pi \epsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}
\]

Here:

- \(\mathbf{r}\) is the vector from the dipole to the point where the potential is being calculated.
- \(\mathbf{p} \cdot \mathbf{r}\) is the dot product of the dipole moment vector \(\mathbf{p}\) and the position vector \(\mathbf{r}\).

This equation is valid for points far away from the dipole where the approximation of the dipole moment as a single entity (rather than two separate charges) holds true.

### Summary

The expression:

\[
\Phi_{\text{dipole}}(\mathbf{r}) = \frac{1}{4\pi \epsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}
\]

is the potential due to a dipole at a distance \(r\) from the dipole, where \(\mathbf{r}\) is the vector from the dipole to the point of interest.

### Do you have any questions about the derivation?

#### Related Questions
1. How does the potential change if the dipole is not aligned along the \(x\)-axis?
2. Can you explain the concept of a dipole field and its characteristics?
3. What is the significance of the dipole approximation, and when does it break down?
4. How does the electric field of a dipole differ from the potential?
5. What are some practical examples of dipole fields in physics and engineering?

#### Tip
In problems involving dipoles, always consider whether the point of interest is far enough to apply the dipole approximation. This greatly simplifies calculations and gives a good understanding of the behavior of the electric field and potential.

Explore the derivation of the electric potential \(\Phi_{\text{dipole}}\) due to a dipole in electrostatics, including key formulas and concepts. Learn how to calculate electric potential at a distance \(r\) from the dipole using the dipole moment \(\mathbf{p}\) and position vector \(\mathbf{r}\).

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