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This is a theoretical question for which i would like to know an answer with an example.

I'd like to know if its possible to create a setup where the electric field at a point $P$ is $n^{th}$ derivative (for some number $n$) of the electric (or magnetic) field at some other point $Q$.

Assumption : Continuous electric charge distributions exist, which means there is no need to base everything on an electron.

PS: derivative is with respect to space or time.

EDIT : I am adding this in view of comments given below.

Assume that the derivative is with respect to time. My question is some what similar to this situation here, assuming an inductor connected to an ideal current source (AC), then the potential drop across the inductor is proportional to the derivative of the current through it.The only difference here is i need it in terms of fields (assuming any type of charge distributions (except distributions(math))) rather than in a circuit with lumped elements.

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If you want it to work for a particular electromagnetic field, just calculate the $n$th derivative at $Q$, and make sure that the electric field at $P$ is right. If you want to create a machine that makes it true for any field - just make $n+1$ nearby measurements near $Q$, calculate the appropriate derivative, and make the right field at $P$. Do you want an analog gadget that does this job, or something else? It's not quite clear. –  Luboš Motl May 4 '11 at 16:22
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To get you going: have you thought about using multiple differentiator circuits? It's a special example, but should still do the job. –  Gerben May 4 '11 at 16:26
    
@Luboš : I do not understand what is meant by "make sure the electric field at$P$ is right", and also from there on. My question is some what similar to this situation here, assuming an inductor connected to an ideal current source (AC), then the potential drop across the inductor is proportional to the derivative of the current through it.The only difference here is i need it in terms of fields (assuming any type of charge distributions (except distributions(math))) rather than in a circuit with lumped elements. –  Rajesh D May 4 '11 at 16:33
    
Are you asking about Green's functions? en.wikipedia.org/wiki/Green's_function –  Jerry Schirmer May 4 '11 at 18:26
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I totally don't understand the question. The electric field at some point CAN'T be equal to the derivative at some other point: a field and its derivative have different units. Do you mean proportional to its derivative (at some other point)? If that's the case, and you're really only referring to two points and a single value of n, then pretty much ANY distribution would satisfy it. What am I missing here? –  Anonymous Coward May 10 '11 at 16:27
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3 Answers 3

up vote 0 down vote accepted

Here we will consider a single time derivative for simplicity. For technical reasons we will also assume that the band of frequencies $\omega$ for the time-varying electric fields $E_1(t)$ and $E_2(t)$, i.e, the support of the Fourier transforms $\hat{E}_1(\omega)$ and $\hat{E}_2(\omega)$, are restricted within certain limits $\omega_1 \ll \omega \ll \omega_2$.

The idea is now to use a power supply with a time-varying voltage $V(t)$ and two RC circuits, where each capacitor is a parallel-plate capacitor with electric field $E_k=\frac{V_k}{d_k}$, where $d_k$ is the gab distance of the plates, $k=1,2$.

     _________________________
     |           |           |
     <           |           <
     >  R_1      |           >  R_2
     <          --- V(t)     <    
     |           -           |
V_1 === C_1      |      V_2 === C_2
     |           |           |
     |___________|___________|

It follows from, e.g., Kirchhoff's and Ohm's laws that

$$ \frac{V-V_k}{R_k}= I_k = \frac{dQ_k}{dt}= C_k \frac{dV_k}{dt}, \qquad k=1,2,$$

or

$$ V(t) = \left(1 + R_k C_k \frac{d}{dt}\right) V_k(t), \qquad k=1,2.$$

Fourier transformation yields

$$ \hat{V}(\omega) = \int dt \ e^{-i\omega t} V(t) = \left(1 + R_k C_k i\omega \right) \hat{V}_k(\omega), \qquad k=1,2.$$

Now adjust the four components $R_k$ and $C_k$, $k=1,2$, such that

$$\omega_k=\frac{1}{ R_k C_k}, \qquad k=1,2.$$

Then

$$\hat{V}_2(\omega) \approx \hat{V}(\omega) \approx i\frac{\omega}{\omega_1} \hat{V}_1(\omega) \qquad\mathrm{for} \qquad \omega_1\ll \omega \ll \omega_2. $$

By inverse Fourier transformation,

$$V_2(t) \approx V(t) \approx \frac{1}{\omega_1} \frac{dV_1(t)}{dt},$$

and hence, we achieve:

The electric field $E_2(t)=\frac{V_2(t)}{d_2}$ in parallel-plate capacitor 2 is proportional to the time derivative of the electric field $E_1(t)=\frac{V_1(t)}{d_1}$ in parallel-plate capacitor 1.

It should be emphasized that $V(t)\approx V_2(t)$ is the input, so the circuit is actually an integrator rather than a differentiator.

Finally, let us mention that the $n$'th (anti)derivative for instance can be build by appropriately connecting $n$ of such modules, perhaps with $n-1$ amplifiers in-between.

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The electric field of a dipole is the derivatve of the monopole field: to put it more accurately, if you create a dipole with a z orientation, the field is obviously d/dz of the field of a monopole. So if you have a given charge distribution, you just replace all the charges with dipoles and the resulting field is the derivative (with respect to z) of the original field.

I don't think this is what you were asking though.

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For a harmonic field like $E\cdot cos(\omega t - \vec{k}\vec{r})$ is is possible since the second time and space derivatives are proportional to each other. For exponentially depending field (decreasing or increasing) it is also possible. No for any other case.

But it is not calculated as an $n^{th}$ derivative, of course.

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