# Shape of field line in an electric dipole

• darkrystal
In summary, Gauss' law states that the flux through a surface passing through a charged region is only through the bottom of the surface. To find the flux at a point C(m,0), one must integrate over x and then solve for k.
darkrystal

## Homework Statement

In a xy-plane, we have a +q charge in A(0,h) and a -q charge in B(0,-h).
Let's take a field line passing through A, that has horizontal slope in A. Find the point C(m,0) where this field line intersects the X axis.

## Homework Equations

I think Gauss' law $$\epsilon_0 \oint \vec{E} \cdot d\vec{A}=q$$

## The Attempt at a Solution

Let's take a surface passing through A and following the field lines till the y=0 plane. Apply the Gauss' law to it: the flux is only through the base of this surface, so that we should get
$$\displaystyle q=\epsilon_0 \int d\Phi = \epsilon_0 \int_0^m \frac{2khq}{(x^2+h^2)^{\frac32}} \cdot 2 \pi x dx$$​
where the second integral is the flux calculated for concentric rings around the origin. This equation, though, has no solutions for m, since it reduces to
$$\displaystyle h(\frac{1}{h}-\frac{1}{\sqrt{h^2+m^2}})=1$$​
which is clearly impossible...

Thank you all for your time, and please forgive my bad english and my even worse physics :)

It's early and I think I'm missing something here but I'll get the ball rolling, I'm starting by assuming you got to the integral correctly and had the correct integrand

First, what happened to k? Second, you're integrating over x in that integral, the only dependency on x is just, well, that single x

I'll write it in a bit more detail. Let's take the flux across a ring ($$x$$ to $$x+dx$$). The distance of every point of this ring from the two charges is the same, so the electric field must be the same on the whole ring. We get this field by multiplying by 2 the y-component of the field generated by the first charge only (this should be true by symmetry). Putting all together, the field SHOULD be $$(2) \cdot \frac{kq}{x^2+h^2} \cdot \frac{h}{\sqrt{x^2+h^2}}$$, where the first 2 is to take in account the doubling I was talking about, the first fraction is the field generated by the first charge, and the second the cosine of the angle (to get only one the y-component). Then I'm multiplying all by the area of the ring, and that's all for the integrand.
For k... that's what puzzles me, mainly. Taking what I can out of the integral, I get a $$\epsilon_0 4 \pi k$$, and everything magically vanishes...

For the integral, I got $$\int \frac{x}{(x^2+h^2)^{\frac32}} = - \frac{1}{\sqrt{h^2+x^2}}$$, and I think this should be correct (anyways, if anybody is willing to check...)

Just one last thing: in the integrand... there isn't only one x... I couldn't get what you meant, sorry

## 1. What is an electric dipole?

An electric dipole is a pair of equal and opposite electric charges separated by a small distance. In other words, it is a system with a positive charge and a negative charge, which creates an electric field around it.

## 2. How are the shape of field lines in an electric dipole determined?

The shape of field lines in an electric dipole is determined by the orientation and distance between the two charges. The field lines always start from the positive charge and end at the negative charge, forming a U-shaped pattern.

## 3. What is the significance of the shape of field lines in an electric dipole?

The shape of field lines in an electric dipole represents the strength and direction of the electric field. The closer the field lines are to each other, the stronger the electric field is at that point. Additionally, the direction of the field lines indicates the direction in which a positive test charge would move.

## 4. Do the field lines of an electric dipole ever intersect?

No, the field lines of an electric dipole never intersect. This is because intersecting field lines would imply that the electric field has two different strengths and directions at the same point, which is not possible.

## 5. How does the shape of field lines change as the distance between the two charges in an electric dipole increases?

As the distance between the two charges in an electric dipole increases, the shape of the field lines becomes more elongated and the field becomes weaker. This is because the electric field strength is inversely proportional to the distance squared.

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