# Calculate electric field strength from voltage?

1. ### rsalmon

12
Hi,

I am trying to work out the electric field strength associated with a number of different electronic systems at a given distance. I am able to simplify the systems so that I only need worry about the fields from a high-voltage busbar or cable.
I know that electric field strength from a cylindrical wire can be calculated using Gauss' law:

E(R)=$$\lambda$$/(2*Pi*epsilon(0)*R),

lambda = charge per unit length, R = radial distance. (sorry about format - couldn't work out the formula writing)
I know the Voltages, and currents, of my various cables but how can I convert them into charge per unit length?

Rob

2. ### rsalmon

12
Ah I have been a bit stupid! No wonder I couldn't get much help from the internet as it was so obvious!
Electric field strength is in Volts/metre, so I simply divide the voltage value from the cable by the radial distance.
if this is wrong please let me know.

3. ### Born2bwire

1,776
In electrostatics, the voltage is the negative integral of the electric field over a line between your test points. If you want to be exact about it you need to consider the system as a whole and the fact that you have wires of non-zero radius (otherwise you will have a non-integrable singularity at R=0). This has been worked for most common systems I'm sure. You should be able to find the appropriate analysis of common systems like a microstrip line or twisted pair.

4. ### rsalmon

12
Ok well I have had a look around and wasn't able to actually find an example of the problem I have. Also I am not sure about my earlier understanding.
If I have a cable at, say, 1 kV and 1kA and I stand 1m away radially what is the magnetic field strength that I would feel?

My previous understnading was that I would then have 1000V / 1m and thus 1000 V/m electric field strength. is this right?

Is this not the same as saying V = E * d
where d is distance between the two points.

5. ### AJ Bentley

664
What is it you are trying to work out? The magnetic field or the electric field?

The magnetic field (close to the wire) can be calculated quite easily using Ampere's law.

The electric field is not so easy because you are not talking here about the simple case of an isolated, infinitely long, charged conductor. You have a complex arrangement involving a fairly short wire and various other parts which also create a field.

6. ### rsalmon

12
Ah sorry about the confusion. I am trying to to calculate the electric field. I understand that I have got a complicated system which is why, as a physicist, I am attempting to model it as simply as possible.

If I propose another example. A high voltage power line on the UK supergrid, 400kV, standing 10m from the ground. If I were to model it as a single infinately long cable what is the electric field strength on the ground?

7. ### AJ Bentley

664
That's a tricky one.

For a charged wire, in theory the potential falls off exponentially but to calculate actual values you need to start from the radius of the wire.
Then it's only going to be (semi-) accurate close to the wire in a real situation.

For your power cable, the electric field strength (and potential) will be zero at the ground by definition.

{EDIT}
Here's the calculation for the potential (voltage) in the space between two co-axial conductors at some point r between them. Take the outer conductor as 0V and the inner as Vo.

V = Vo * Log(r/b)/Log(a/b)

where a and b are the radii of the inner and outer conductors. (natural logs of course)

You can generalise this roughly to a long, high voltage wire if a is the wire radius and b is the distance to ground in the direction you are interested in. It will be a very approximate answer.

Last edited: Aug 23, 2010
8. ### Born2bwire

1,776
V = -Ed actually and it only works if the electric field is uniform and constant. In your case, you are looking at an electric field that varies in space and thus you need to take the line integral to find the potential difference.

Now in case of your high wire example, we can model the ground to a roughly as a conductor. This means that we can use image theory and thus if you have a charged wire a distance 10 m above the ground then we can model the entire system as having a second oppositely charged wire 10 m below the ground. So we simply have two infinite parallel wires with opposite charges. You can then find the electric field for a given unknown charge density (via the equation you gave in your original post) using superposition. Then you could integrate between the midpoint between the two wires and to the surface of the wire (again you need to have a non-zero radius otherwise it will be singular) and then you find the voltage. Solve for the charge density given your original voltage (400 KV) and then substitute back into the electric field equation to find the actual fields.

Of course though, for the specific question of the electric field strength at the ground it will be zero.

But I have a feeling that there are a myriad of books that deal with the above problems. You should be able to do a literature search through power engineering or electrical engineering (look at transmission lines) texts and find many of these basic examples worked out for you. Also note again that the above analysis changes when we talk about AC voltages as then we need to start using electromagnetic wave theory.

9. ### rsalmon

12
Ok, Well I have been trawling the internet and the the books again looking for examples of this problem I am having. Unfortunately I am still in confusion.

I'm unsure what it is that I need to intergrate (what equation I mean) would you be able to clarify.

I didn't quite understand the method here. Do I use my initial equation and use a guessed charge density and then ..... I'm sorry I'm just a little confused. perhaps a few equations would help clarfiy.

I have discovered that if it is an alternating current then a tim-varying magnetic field will be produced and thus by faradays law ( curl E = dB/dt) cause an induced magnetic field. I assume that is as well as a normal electric field generated by the charges in the wire. For the time being I would just like to know how to do a DC analysis.

10. ### vanhees71

3,815
For static situations the Maxwell equations split into two pairs, the electric-field equations decouple from the magnetic-field equations. In the usual macroscopic electromagnetics they read (in Heaviside-Lorentz units)

Maxwell equations (statics)

$$\vec{\nabla} \times \vec{E}=0,$$ (1)
$$\vec{\nabla} \cdot \vec{D}=\rho,$$ (2)
$$\vec{\nabla} \cdot \vec{B} = 0,$$ (3)
$$\vec{\nabla} \times \vec{H}=\frac{1}{c} \vec{j},$$ (4)

Constitutive Equations (homogeneous isotropic media)

$$\vec{D}=\epsilon \vec{E}$$ (5)
$$\vec{B}=\mu \vec{H}$$ (6)
$$\vec{j}=\sigma \vec{E}$$ (7)

Concentrating on electrostatics, we need only Eqs. (1), (2), and (5).

Eq. (1) tells you that the electric field is a gradient field (in simply connected regions in space)

$$\vec{E}=-\vec{\nabla} \Phi$$ (8)

and through equation (6) you get (for $$\epsilon=\text{const}$$)

$$\Delta \Phi=-\frac{1}{\epsilon} \rho$$.

For a given charge distribution the solution, if there are not any boundary conditions to consider, is given by the Green's function of the Laplace operator

$$\Phi(\vec{x})=\int_{\R^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi \epsilon |\vec{x}-\vec{x}'|}$$

Physically that's easy to understand: you just add the potentials for each charge-volume element given by Coulomb's law, leading to the integral in the limit of arbitrarilly small volume elements.

For a given given potential, you get the electric field by the gradient, cf. Eq. (8). If you like to evaluate the potential for a rotation free electric field, obeying Eq. (1), you either solve the corresponding first-order system of partial differential equations or you use the line-integral representation

$$\Phi(\vec{x})=-\int_{C(\vec{x},\vec{x}_0)} \mathrm{d} \vec{x'} \cdot \vec{E}(\vec{x}'),$$

where $$C(\vec{x},\vec{x}_0)$$ is an arbitrary path connecting the ponts $$\vec{x}_0$$ with $$\vec{x}$$ within a simply connected region in space, where $$\vec{E}$$ is well defined and curl free. Then the integral is also independent of the particular path connecting these two points, and the potential is thus uniquely defined within this simply connected region.

11. ### rsalmon

12
Hi,

Thanks for all the posts. I have now got a formula of sorts that has a more simplified approach along the same lines that has been indicated above.

Problem; power line (0.05m radius) 12m high operating at 400kV DC. Observer is 1m from the ground. what is the electric field strength felt by the observer?

Method:
If I model it similiar to a coaxial cable where the observer is at the position of the outer conductor. I take the voltage across as the 400kV and so can use the following equation;

V = - int[q/(2*Pi*epsilon{0}*r]dr --(can't get the equation editor to work)

where the limits are A (inner radius of wire) and B (outer radius). so:

V = q/(2*Pi*epsilon{0}) *ln(B/A)
q=(2*Pi*epsilon{0}*V)/ln(B/A)

I also have;

E = q/(2*Pi*epsilon{0})

leaving me with;

E=V/[ln(B/A)*r]

so for B=11m A=0.05m r=11m and V=400kV
E=6.74 kV/m

Am I getting close or is this far too simple / wrong application of formula? (I apologise for the formatting).

P.S. There is a website that shows the answers to this problem but not the full method. (after asking them directly they just use a computer program)

12. ### Don Kelly

4
Sorry to be so late to this thread. a previous person (born...) has it right. If you consider the use of an image 12m below ground, the ground plane will be an equipotential which is not concentric with the conductor as in a coax.
400=(1/2pie0)rho* ln2h/r as a result where 2h=2*12m and r =0.05m
This gives the charge.
Now E=(rho/2pie0)(1/r +1/r') where r and r' are the distances between the conductor and the point under consideration and r' is the same for image to point.

rho =6.174 (2pie0) kC/m
so at 1m above ground directly below the conductor, the field is vertical at
6.174(1/11 +1/13)=1.04 Kv/m downward This is close enough to the ground level field that the ground level field of 1.03KV/m can be used. Even at 6m, the field is only 1.4KV/m
At the surface of the conductor it is 123KV/m
By the way a 5cm radius conductor is mechanically impractical, generally a bundle of smaller conductors is used- say 3 1cm radius conductors spaced 20 to 30 cm apart.

This electrostatic approach can be extended to multi conductor lines and also for 50-60Hz AC power lines as the frequency is low. Full fledged field theory isn't needed.