- #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
- 11,844
- 13,536
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
- 15,012
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:
1. What is a dipole?
A dipole is a pair of equal and opposite charges separated by a distance.
2. How is the electric field due to a dipole calculated?
The electric field due to a dipole can be calculated using the formula: E = k * (p / r^3), where E is the electric field, k is the Coulomb constant, p is the dipole moment, and r is the distance from the dipole.
3. In which direction does the electric field point for a dipole?
The electric field points from the positive charge to the negative charge of the dipole.
4. What happens to the electric field as you move farther away from the dipole?
As you move farther away from the dipole, the strength of the electric field decreases inversely with the cube of the distance.
5. Can a dipole have a net electric field at a certain point?
Yes, a dipole can have a net electric field at a certain point if the distance from the dipole is large compared to the separation between the charges.
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