# Question with simple current and charge of an electric dipole.

• yungman
In summary: Icos(wt). So they cancel out.In summary, the book's equation for the current and charge in a hertzian dipole is incorrect because Q is imaginary.
yungman
This is from Field and Waves Electromagnetic by Cheng page 602. THis is regarding to elemental dipoles. Given: [Quote Book]
$$i(t)=\Re e [Ie^{j\omega t}]=I\; cos\;\omega t\;\;\hbox { (11.7)}$$

Since the current vanishes at the ends of the wire, charge must be deposited there.

$$i(t)=^+_-\frac {d q(t)}{dt} \;\;\hbox { (11.8)}$$

In phasor notation, $q(t)=\Re e[Qe^{j\omega t}]$, we have

$$I=^+_-j\omega Q\;\;\hbox{ (11.9)}$$

or $$Q = ^+ _- \frac {I} {j \omega}\;\;\hbox {(11.10)}$$

[End Book]

This is what my calculation of q(t):

$$i(t)=\frac {d [q(t)]}{dt} = \Re e \left [ Q \frac {d (e^{j\omega t})}{dt}\right ] =\Re e [ Q j\omega e^{j\omega t}] = -\omega \;Q \;sin\;\omega t$$

You see $i(t)$ not equal to $I\; cos\;\omega t$

Last edited:
But you defined I to be a complex number and then you relate the real valued i(t) with I\cos \omega t which is complex.

I did not define anything. The first part I copy from the book. Looks to me both I and Q are a constant which stand for the peak value of the current and charge resp. as you see the $cos\;\omega\;t$ that make the current/charge harmonically time varying.

I double check the differentiation in phasor form and showed the result of differentiating q(t) resp to t and it don't show $I\;cos\;\omega t$

Thanks

yungman said:
I did not define anything. The first part I copy from the book. Looks to me both I and Q are a constant which stand for the peak value of the current and charge resp. as you see the $cos\;\omega\;t$ that make the current/charge harmonically time varying.

I double check the differentiation in phasor form and showed the result of differentiating q(t) resp to t and it don't show $I\;cos\;\omega t$

The book did something wrong then. As Born2bwire mentioned, you have this formula:

$$I=\pm j \omega Q$$
If Q is real, then I is complex.

If I is complex, then this formula does not hold:

i(t)=ℜe[Iejωt]=Icosωt

If Q is not real, then this formula does not hold:

q(t)=ℜe[Qejωt]=Qcosωt

So there are some inconsistencies in your equations.

Thanks for the help. I edit the original post. I actually type the paragraph word to word straight from the book and then at the bottom I show my evaluation that show the inconsistency.

I know, something seems wrong with the book and I cannot make sense out of it.

Given

$$i(t)=\Re e [Ie^{j\omega t}]=I\; cos\;\omega t\;\;\hbox { (11.7) , }q(t)=\Re e[Qe^{j\omega t}]\;\;\;\hbox { and }\;\;I=^+_-j\omega Q\;\;\hbox{ (11.9)}$$

You really lock it in. I hope by typing word for word, you might make some sense out of it.

Thanks

Alan

yungman said:
Thanks for the help. I edit the original post. I actually type the paragraph word to word straight from the book and then at the bottom I show my evaluation that show the inconsistency.

I know, something seems wrong with the book and I cannot make sense out of it.

Given

$$i(t)=\Re e [Ie^{j\omega t}]=I\; cos\;\omega t\;\;\hbox { (11.7) , }q(t)=\Re e[Qe^{j\omega t}]\;\;\;\hbox { and }\;\;I=^+_-j\omega Q\;\;\hbox{ (11.9)}$$

You really lock it in. I hope by typing word for word, you might make some sense out of it.

Thanks

Alan

You derived this:

i(t)=d[q(t)]/dt=ℜe[Qd(ejωt)/dt]=ℜe[Qjωejωt]=−ωQsinωt

which is incorrect.

Q is imaginary, so you can't just take it out of the Re[]. What you do is plug in Q=+- I/(jw), and then take the real part. Then you should get something like: I(t)=Icos(wt)

RedX said:
You derived this:

i(t)=d[q(t)]/dt=ℜe[Qd(ejωt)/dt]=ℜe[Qjωejωt]=−ωQsinωt

which is incorrect.

Q is imaginary, so you can't just take it out of the Re[]. What you do is plug in Q=+- I/(jw), and then take the real part. Then you should get something like: I(t)=Icos(wt)

This is the hertzian dipole, I don't get why Q is imaginary. Convensionally, Q is just the amplitude and q(t) is just Q cos wt.

yungman said:
This is the hertzian dipole, I don't get why Q is imaginary. Convensionally, Q is just the amplitude and q(t) is just Q cos wt.

Your equation (11.9) says that one of the quantities must be imaginary. So it's either Q or I.

A Hertzian dipole behaves like a capacitor, where the current leads the voltage. Now for a capacatitor Q follows V. So you expect the current to lead the charge. Since a hertzian dipole behaves like a pure capacitance, the lead is 90 degrees.

So if your current behaves like cos(wt), then the charge would lag behind by 90, or cos(wt-90)=sin(wt).

So both the charge and voltage can't both vary as cos(wt). One of them has to vary as sin(wt).

So notice Q=Real[Qe(iwt)]=Real[I/(jw)e(iwt)]=(I/w)sin(wt)

So Q varies by sin(wt), and I by cos(wt).

Last edited:
Thanks, I think I got it.

Good night.

RedX said:
Your equation (11.9) says that one of the quantities must be imaginary. So it's either Q or I.

A Hertzian dipole behaves like a capacitor, where the current leads the voltage. Now for a capacatitor Q follows V(Do you mean I, it should not be V). So you expect the current to lead the charge. Since a hertzian dipole behaves like a pure capacitance, the lead is 90 degrees.

So if your current behaves like cos(wt), then the charge would lag behind by 90, or cos(wt-90)=sin(wt).

So both the charge and voltage can't both vary as cos(wt). One of them has to vary as sin(wt).

So notice Q=Real[Qe(iwt)]=Real[I/(jw)e(iwt)]=(I/w)sin(wt)

So Q varies by sin(wt), and I by cos(wt).
Please see the question in Blue above.

yungman said:
Please see the question in Blue above.

I meant voltage. For a capacitor, Q=CV, where C (the capacitance) is real. So whatever phase the voltage has, the charge Q has the same phase. A Hertzian dipole (i.e. a very tiny antenna) is pretty much an AC generator connected to a capacitor. If you want you can put two spheres at the end of the wires coming out of the generator, so it looks more like a capacitor. In a circuit that contains just a capacitor and a generator, if the current is I(t)=I cos(wt) where I is real, then the voltage is V(t)=I/(wC) sin(wt). The charge would then be C times this, or Q(t)=(I/w)sin(wt).

This is the same result that you got. The charge varies as sin(wt), and the current as cos(wt), so the charge and current aren't in phase but are 90 degrees out of phase.

This capacitor analogy only works if the dipole is very tiny (compared to the wavelength of the frequency that it's being driven at). As the dipole gets bigger the behavior of the antenna changes, sometimes acting as pure inductors, or series and parallel resonant circuits.

So basically I was trying to give another picture of what's going on in the antenna. You can model it as an AC generator connected to a pure capacitor, and solve for the charge and current as you would a normal circuit.

I mis-understood. When I saw "Q follow V", I thought you meant Q is behind V...V first and then Q like Q follow ( lagging behind) I in phase. I understand Q is in phase with V where Q=CV.

Thanks

Alan

## 1. What is an electric dipole?

An electric dipole is a pair of equal and oppositely charged particles, separated by a small distance. It can also refer to a molecule with a slightly positive end and a slightly negative end due to differences in electron distribution.

## 2. How is the current in an electric dipole calculated?

The current in an electric dipole is equal to the product of the magnitude of the charges and the distance between them, divided by the time it takes for the charges to move. This can be represented by the formula I = Qd/t, where I is the current, Q is the charge, d is the distance, and t is the time.

## 3. What is the direction of the current in an electric dipole?

The direction of the current in an electric dipole is from the positive charge to the negative charge. This is because the positive charge has a higher potential energy and will move towards the negative charge, creating a flow of charge.

## 4. How does the charge of an electric dipole affect its behavior?

The charge of an electric dipole affects its behavior in several ways. Firstly, a stronger charge will result in a stronger current, as seen in the formula I = Qd/t. Additionally, the larger the charge, the stronger the electric field around the dipole will be, which can influence the movement of other nearby charges.

## 5. Can an electric dipole have a net charge of 0?

Yes, an electric dipole can have a net charge of 0 if the magnitude of the charges is equal and they are located at the same distance from their center. In this case, the electric fields created by the two charges cancel each other out, resulting in a net charge of 0 for the dipole.

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