Electric Field Intensity

In summary, the conversation discusses the independence of the intensity of the electric field between two parallel plates on the separation distance and its dependence only on the charge on the plates. It is mentioned that for a parallel plate capacitor with plate separation much smaller than the plate diameter, the field is constant and can be expressed using Gauss's law or the equation E = V/d. However, when the plates are connected to a battery, the field becomes dependent on the distance between the plates. The concept of symmetry and the use of a single plate are also discussed.
  • #1
I have a text that claims that the intensity of the electric field between two parallel plates is independent of the separation distance and depends only on the charge on the plates. Is this correct?
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  • #2
Only so long as the plate diameter (or, for a square plate, side length) is much larger than the plate separation.
  • #3
As a general statement it sounds wrong to me !
The equation for electric field STRENGTH (I assume that is what your text means as intensity) is
E = V/d where V = the voltage between the plates and d is their separation.
Where did the text come from?
  • #4
Your'e right, technician, that E = V/d, but for the parallel plate capacitor with plate separation << plate diameter, Gauss's law yields another expression for E, namely E = Q/[(epsilon-nought)A]. So if (like David Furlong) we use charge rather than voltage as our electrical variable, E is indeed independent of plate separation, with the proviso I gave above.
  • #5
Assuming a constant charge (capacitor disconnected from battery) the voltage will increase when you increase the distance between the plates so that the ratio V/d stays the same and the electric field will be constant. Of course, as long as the distance is small compared with the width of the plates.
  • #6
Thanks for responses Nasu and Philip !
I had a battery connected to parallel plates in mind.
  • #7
In this case (battery connected) the electric field will be dependent on the distance between plates. The charge will change when d changes.
  • #8
It's because of symmetry. If the width and height of the plates are much larger than their separation, they are effectively flat and infinite planes. For a flat infinite plane of uniform charge distribution, symmetry dictates that the only possible field produced is uniform and constant. And you don't need two plates. One plate will have the same result. If you are very close to a single plate with uniform charge, such that it appears to be infinite and height and width, the field will be constant and uniform.

1. What is electric field intensity?

Electric field intensity, also known as electric field strength, is a measure of the strength of an electric field at a given point in space. It is defined as the force per unit charge experienced by a small test charge placed at that point.

2. How is electric field intensity calculated?

Electric field intensity is calculated by dividing the force experienced by a test charge by the magnitude of the charge. This can be mathematically represented as E = F/Q, where E is the electric field intensity, F is the force, and Q is the charge.

3. What are the SI units of electric field intensity?

The SI units of electric field intensity are newtons per coulomb (N/C) or volts per meter (V/m).

4. How does distance affect electric field intensity?

According to Coulomb's Law, electric field intensity is inversely proportional to the square of the distance between the charges. This means that as the distance between two charges increases, the electric field intensity decreases.

5. What are some real-life applications of electric field intensity?

Electric field intensity is an important concept in many fields, including electrical engineering, physics, and chemistry. It is used in the design of electronic circuits, the study of lightning and atmospheric electricity, and in medical imaging techniques such as MRI and EKG.

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