- #1
wolly
- 49
- 2
and
Can someone explain how they made these equations like this?
How did the radius become that equation?
What formulas from algebra did they applied?
I'm looking at these formulas and I don't understand how r=z+1/2*d
The electric field due to a dipole
- I
- Thread starter wolly
- Start date
and
Can someone explain how they made these equations like this?
How did the radius become that equation?
What formulas from algebra did they applied?
I'm looking at these formulas and I don't understand how r=z+1/2*d
- #2
Ibix
Science Advisor
- 12,227
- 14,089
The dipole charges are defined to be on the z axis at ##z=\pm d/2##. If I point out that your textbook is not using full-on vector algebra, so is working in 1d, can you see how to get from ##r^2## to ##(z\pm d/2)^2##?
- #3
- 24,492
- 15,025
From this little example, I can only come to the conclusion to recommend to use another book. SCNR.
The idea of the dipole is to have a charge ##Q## at ##(0,0,d/2)=d/2 \vec{e}_3## and ##-Q## at ##(0,0,-d/2)=-d/2 \vec{e}_3##. The field is given by the superposition of the field of these point charges (Coulomb fields), i.e.,
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0} \left [\frac{\vec{r}-d/2 \vec{e}_3}{|(\vec{r}-d/2 \vec{e}_3|^3}-\frac{\vec{r}+d/2 \vec{e}_3}{|\vec{r}+d/2 \vec{e}_3|^3} \right].$$
The next step usually is to make ##d \rightarrow 0## while keeping ##Q d=\text{const}##. This leads to the dipole approximation of the field for this charge configuration. This is a somewhat cumbersome calculation, however.
It's easier to use the electrostatic potential:
$$\Phi(\vec{r},d)=\frac{Q}{4 \pi \epsilon_0} \left [\frac{1}{|\vec{r}-d/2 \vec{e}_3|} - \frac{1}{|\vec{r}+d/2 \vec{e}_3|} \right].$$
Now ##|\vec{r} \pm d/2 \vec{e}_3|=\sqrt{x_1^2 +x_2^2 + (x_3\pm d/2)}^2##.
The Taylor expansion of ##\Phi## wrt. ##d## is
$$\Phi(\vec{r},d)=\Phi(\vec{r},0) + d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2) = d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2).$$
This gives
$$\Phi(\vec{r},d)=\frac{Q d x_3}{4 \pi \epsilon_0 r^3} + \mathcal{O}(Q d^2).$$
Defining the dipole moment
$$\vec{p}=Q d \vec{e}_3$$
Making now ##d \rightarrow 0## but keeping ##Q d=\text{const}## you get the dipole potential,
$$\Phi(\vec{r})=\frac{\vec{p} \cdot \vec{r}}{4 \pi \epsilon r^3}.$$
The field is
$$\vec{E}=-\vec{\nabla} \Phi=-\frac{\vec{p}}{4 \pi \epsilon_0 r^3} + \frac{3 (\vec{p} \cdot \vec{r}) \vec{r}}{4 \pi \epsilon_0 r^5} = \frac{1}{4 \pi \epsilon_0 r^5} [3 (\vec{r} \cdot \vec{p}) \vec{r}-r^2 \vec{p}].$$
The idea of the dipole is to have a charge ##Q## at ##(0,0,d/2)=d/2 \vec{e}_3## and ##-Q## at ##(0,0,-d/2)=-d/2 \vec{e}_3##. The field is given by the superposition of the field of these point charges (Coulomb fields), i.e.,
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0} \left [\frac{\vec{r}-d/2 \vec{e}_3}{|(\vec{r}-d/2 \vec{e}_3|^3}-\frac{\vec{r}+d/2 \vec{e}_3}{|\vec{r}+d/2 \vec{e}_3|^3} \right].$$
The next step usually is to make ##d \rightarrow 0## while keeping ##Q d=\text{const}##. This leads to the dipole approximation of the field for this charge configuration. This is a somewhat cumbersome calculation, however.
It's easier to use the electrostatic potential:
$$\Phi(\vec{r},d)=\frac{Q}{4 \pi \epsilon_0} \left [\frac{1}{|\vec{r}-d/2 \vec{e}_3|} - \frac{1}{|\vec{r}+d/2 \vec{e}_3|} \right].$$
Now ##|\vec{r} \pm d/2 \vec{e}_3|=\sqrt{x_1^2 +x_2^2 + (x_3\pm d/2)}^2##.
The Taylor expansion of ##\Phi## wrt. ##d## is
$$\Phi(\vec{r},d)=\Phi(\vec{r},0) + d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2) = d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2).$$
This gives
$$\Phi(\vec{r},d)=\frac{Q d x_3}{4 \pi \epsilon_0 r^3} + \mathcal{O}(Q d^2).$$
Defining the dipole moment
$$\vec{p}=Q d \vec{e}_3$$
Making now ##d \rightarrow 0## but keeping ##Q d=\text{const}## you get the dipole potential,
$$\Phi(\vec{r})=\frac{\vec{p} \cdot \vec{r}}{4 \pi \epsilon r^3}.$$
The field is
$$\vec{E}=-\vec{\nabla} \Phi=-\frac{\vec{p}}{4 \pi \epsilon_0 r^3} + \frac{3 (\vec{p} \cdot \vec{r}) \vec{r}}{4 \pi \epsilon_0 r^5} = \frac{1}{4 \pi \epsilon_0 r^5} [3 (\vec{r} \cdot \vec{p}) \vec{r}-r^2 \vec{p}].$$
Last edited:
Related to The electric field due to a dipole
What is an electric dipole?
An electric dipole consists of two equal and opposite charges separated by a small distance. It is characterized by its dipole moment, which is a vector quantity pointing from the negative charge to the positive charge.
How do you calculate the electric field due to a dipole at a point along the axial line?
The electric field due to a dipole at a point along the axial line (the line extending through both charges) can be calculated using the formula: \( E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3} \), where \( p \) is the dipole moment and \( r \) is the distance from the center of the dipole to the point.
How do you calculate the electric field due to a dipole at a point along the equatorial line?
The electric field due to a dipole at a point along the equatorial line (the line perpendicular to the dipole axis and passing through the midpoint of the dipole) can be calculated using the formula: \( E = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3} \), where \( p \) is the dipole moment and \( r \) is the distance from the center of the dipole to the point.
What is the direction of the electric field due to a dipole?
The direction of the electric field due to a dipole varies depending on the position relative to the dipole. Along the axial line, the field points away from the positive charge and towards the negative charge. Along the equatorial line, the field points from the positive charge to the negative charge, perpendicular to the dipole axis.
How does the electric field due to a dipole vary with distance?
The electric field due to a dipole decreases with the cube of the distance from the dipole. Specifically, it varies as \( \frac{1}{r^3} \), where \( r \) is the distance from the center of the dipole. This rapid decrease with distance is a characteristic feature of dipole fields.
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