# Homework Help: Electric fields of infinite line & electric dipole

1. Feb 4, 2007

### kingwinner

I am having some problem with the formulas for calculating the electric fields of an infinite line of charge and an electric dipole. I don't understand conceptually why they are the way they are. Can someone explain? Any help is appreciated!

[Note: K=1/(4*pi*epsilon_o), lambda=linear charge density, p=electric dipole moment, E=electic field, r is the distance]

Point charge:
E=Kq/r^2 <---the electric field falls off as 1/r^2.

Infinite line of charge:
E=K(2 lambda)/r <---the field falls off as 1/r, i.e. falls off slower than that of a point charge as you move further away.

Electric dipole:
E=K(2p)/r^3 (field on axis)
E=-Kp/r^3 (field in bisecting plane)
<---the field falls off as 1/r^3, i.e. falls off faster than that of a point charge as you move further away.

Now, is there any easy way to explain the distance-dependence of the electric field of the electric dipole and infinite line of charge? (the colored part above) Why do the formulas make sense? What actually determines the rate at which the electric field falls off as you move away from the charge?

Thank you for explaining!:)

2. Feb 4, 2007

### mjsd

firstly, you have to believe in Gauss's Law: total electric flux (field x area) through a closed surface is equal to charge enclosed divided by $$\varepsilon_0$$.

Ok, so for the point charge, the natural Gaussian surface to choose is "the sphere" (because of the radial symmetry expected from the field lines generated by the point charge)... and you get
$$E.4\pi r^2 = q_\text{enclosed}/\varepsilon_0$$
rearranging you get the $$1/r^2$$ dependence

for the infinite line, the natural Gaussian surface to choose in this case is "the cylinder" and you have
$$E.A_\text{lateral} + E.A_\text{ends}= E. 2\pi r h + 0 = q_\text{enclosed}/\varepsilon_0$$
second term vanish because E is perpendicular to the normal of the surface. Rearranging, you get the $$1/r$$ dependence.

for dipole case it is a bit more complicated. Your quoted answer is true in the "far field" approximation that is when distance r is a lot bigger than the separation of the charges. for an intuitive feel why it falls off quicker: from a distance the dipole looks like two equal and opposite charges that almost coincide. Thus, at a distance, they almost but not quite cancel each other.
to show this mathematically, just use the formula for a point charge and do this for both charges and ADD them up. the resulting expression is a difference bewteen two expressions of the form $$(1+x)^{-2}$$. Each can then be expanded using the binomial theorem and using the "far field" approximation (r > d, where dipole moment: p = qd), you get your formula.

3. Feb 4, 2007

### kingwinner

For the infinite line of charge, is there any intuitive way to justify why the electric field falls off slower than that of a point charge as you move further away?

Furthermore, how come the electric field of an infinite plane of charge is totally independent of the distance from it? Why does this result make sense intuitively?

4. Feb 4, 2007

### mjsd

I guess Gauss's law is just another way of saying conservation of "stuff"... stuff in this case is probably Electric flux.. or something along those lines... I am being loose with my terminologies here... anyway... field lines carry energy so I guess it is like conservation of "energy"... ok to the "intuitive" view... for a point charge field lines/flux goes in all directions (radial), for the INFINITE line charge (by the way there is no such thing in practice, that's why you may find it hard to picture) , .... now you can imagine a line is just made of a lot of points....but so to prevent violation of conservation laws, some of the field lines are being squashed together making the "field strength" stronger at the same distance r away.... because field lines don't "cross" they can only "emit" perpendicular to the line itself.

In fact this is summarised nicely by Gauss's law.... previously for point charge, the flux must be evenly shared about a $$4\pi r^2$$ surface, now for the line, just $$2\pi r h$$, and since it is infinitely long, h does not come into it at all.

put it simply, for the line, every point element will have another point element next to it to help "reinforce" the field at a perpendicular distance r away from the original point element.. etc.