(a) Write an expression for the magnitude of the electric...

(a) Write an expression for the magnitude of the electric field at a point $(x, 0)$ on a line perpendicular to the dipole axis. State the direction of the field for $x>0$ and for $x<0$ (b) Show that when $x \gg d, E \approx k q d / x^{3}$ (c) The field is inversely proportional to the distance cubed. Does this conflict with Coulomb's law? Explain.

(a) Write an expression for the magnitude of the electric field at a point $(x, 0)$ on a line perpendicular to the dipole axis. State the direction of the field for $x>0$ and for $x<0$
(b) Show that when $x \gg d, E \approx k q d / x^{3}$
(c) The field is inversely proportional to the distance cubed. Does this conflict with Coulomb's law? Explain.

Key Concepts

Electric Dipole Field

This concept entails the electric field produced by a pair of equal and opposite charges separated by a small distance. The field at any point is determined by vectorially adding the contributions from each charge. This involves using Coulomb's law to calculate the magnitude and direction of the field due to individual charges, and then applying the principle of superposition to determine the net field at the point of interest.

Dipole Moment

The dipole moment is a measure of the separation of positive and negative charges in a dipole and is defined as the product of one of the charges and the distance between the charges. It is a key parameter in determining the strength and orientation of the dipole’s electric field. In analyses involving electric dipoles, the dipole moment provides a convenient and compact description of the system’s overall charge distribution.

Far-field Approximation

When examining the electric field at distances much larger than the separation between the dipole charges (x ? d), one can simplify the expression for the field using an approximation. This involves neglecting higher order terms and expressing the leading term in the field expansion, which typically reveals an inverse cubic dependence on distance. This approximation is critical for understanding the behavior of the field in the regime where the spatial variation of the dipole's influence becomes very gradual.

Coulomb's Law and Multipole Expansion

Coulomb's law directly describes the electric field due to point charges, showing an inverse square law dependence on distance. However, when charges are arranged in a dipole configuration, the field results from the combination of two inverse square contributions that, under appropriate geometric considerations and cancellations, yield a net field that decays as the inverse cube of distance. This is an example of a multipole expansion, where the dipole term becomes dominant at large distances and does not conflict with Coulomb's law because it results from the vector superposition of fields from individual charges.

Transcript

00:01 Okay, we have two point charges on the y axis, a distance d apart, and have a positive charge on the positive y axis and a negative charge on the negative y axis.

00:13 Their charges are equal and opposite.

00:18 We're going to call, so we're going to have a point p out here at a distance x along the x axis.

00:25 We have r is actually the same for both charges, is the distance from the charge to the point.

00:32 And of course, we need that to use kulom's law to figure out the electric field.

00:41 We see that, i'm using pythagorean theorem, that r squared is d over two quantity squared plus x squared.

00:51 So the electric field at point p is going to be k times q over r squared times the r1 unit vector.

01:01 We'll get it to that in a minute.

01:04 So the positive charge is charge one, the negative charge is charged one, the negative charge is charge 2.

01:10 So the r1 unit vector points from charge 1 to point p and the r2 unit vector points from charge 2 to point p.

01:27 So the r1 unit vector is the vector from the positive charge out to the point p is going to be x times i minus d over 2 times y or j excuse me.

01:45 And then the r1 unit vector also has a 1 over r factor just like the r2 just like r1 did so those are our two unit vectors that are basically they point from the charge to the point our two points from the negative charge out to the point the minus sign here comes from the on the total field comes from the fact that charge 2 is negative it has the same value as charge 1 so the electric field will factor out the kq over r cubed.

02:32 And we just get some vectors here.

02:34 They come from the unit vectors.

02:39 And then we realize that the x components are equal and opposite and they cancel.

02:47 The y components are the same and we'll add minus two times d over two times j.

03:05 And when r is really large, much larger than d, we have that r is about equal to x.

03:16 And the farther we are out on the x axis, the more likely the more true this becomes...

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