Q about electric field between two parallel plates

In summary, there are two expressions for electric field between two parallel plates, one is E=σ/ε0 and the other is E=V/d. These expressions can be used in different conditions, such as when using Gauss's law or finding the potential difference between two points. They may seem contradictory, but they are just re-arrangements of each other and do not contradict each other. The quantity in this relationship is capacitance, and keeping Q constant while increasing d will require work, resulting in an increase in V but a constant electric field.
  • #1
asdff529
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There are two expressions of electric field between two parallel plates,say one carries Q and another carries -Q
Then the electric field between them=σ/ε0
But there is another expression that E=V/d where d is their distance of separation
What are the differences between them?And what are the conditions when using either of one?
Thank you!
 
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  • #2
Consider Gauss's law [itex] \oint \vec{E}\cdot\hat{n}da=\frac{q}{\varepsilon_0} [/itex].
Now take, as a Gaussian surface, a rectangular cube which includes only part of one of the planes and is far enough from the edges. The charge that it includes is [itex] q=lw\sigma [/itex] where l and w are dimensions of the part of the cube which is parallel to the charged planes.From the symmetry, we know that the electric field is perpendicular to the plane of the cube parallel to the charged planes and is constant all over it and so the surface integral is just [itex] E lw [/itex] and so we have [itex] E=\frac{\sigma}{\varepsilon_0} [/itex].
Now consider [itex] V=-\int_a^b \vec{E}\cdot\vec{dr} [/itex] which is used for finding the potential difference between points a and b from the electric field.If electric field is constant along the way and isn't changing direction,the integral will be just the product of electric field and path length and we will have [itex] E=\frac V d [/itex]. As you saw in the last paragraph,the electric field between the planes was constant along their separation and so the formula [itex] E=\frac V d [/itex] can be used in that case.For example you can have [itex] V=\frac{d\sigma}{\varepsilon_0} [/itex] for the potential difference between two points between the charged planes with separation d.
Its not that they are two different formulas.They're just in terms of different things.
 
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  • #3
Shyan said:
Consider Gauss's law [itex] \oint \vec{E}\cdot\hat{n}da=\frac{q}{\varepsilon_0} [/itex].
Now take, as a Gaussian surface, a rectangular cube which includes only part of one of the planes and is far enough from the edges. The charge that it includes is [itex] q=lw\sigma [/itex] where l and w are dimensions of the part of the cube which is parallel to the charged planes.From the symmetry, we know that the electric field is perpendicular to the plane of the cube parallel to the charged planes and is constant all over it and so the surface integral is just [itex] E lw [/itex] and so we have [itex] E=\frac{\sigma}{\varepsilon_0} [/itex].
Now consider [itex] V=-\int_a^b \vec{E}\cdot\vec{dr} [/itex] which is used for finding the potential difference between points a and b from the electric field.If electric field is constant along the way and isn't changing direction,the integral will be just the product of electric field and path length and we will have [itex] E=Vd [/itex]. As you saw in the last paragraph,the electric field between the planes was constant along their separation and so the formula [itex] E=Vd [/itex] can be used in that case.For example you can have [itex] V=\frac{\sigma}{d\varepsilon_0} [/itex] for the potential difference between two points between the charged planes with separation d.
Its not that they are two different formulas.They're just in terms of different things.

but my teacher said if we use E=σ/ε0 and Q is kept constant,the electric field is independent of d.
it seems that there is a contradiction,because E=V/d as well,where am i wrong?
 
  • #4
asdff529 said:
but my teacher said if we use E=σ/ε0 and Q is kept constant,the electric field is independent of d.
it seems that there is a contradiction,because E=V/d as well,where am i wrong?

There is no contradiction. E=V/d doesn't mean E depends on d! Because V can be a function of d as well.
 
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  • #5
asdff529 said:
but my teacher said if we use E=σ/ε0 and Q is kept constant,the electric field is independent of d.
it seems that there is a contradiction,because E=V/d as well,where am i wrong?
There is no contradiction. The one expression can be re-arranged into the other.
The quantity in this relationship is Capacitance (C) and Q = CV
You can replace this by σ=c0V
where c0 is the capacitance per unit area.

Keeping Q constant and increasing d will require work, so V will have increased. The Volts per Meter will remain the same. Alternatively, separating the plates will decrease the Capacitance, which implies an increase in V.
 

1. What is an electric field between two parallel plates?

The electric field between two parallel plates is a region in space where electric charges experience a force. It is created by placing two charged parallel plates close to each other, with opposite charges on each plate.

2. How is the electric field between two parallel plates calculated?

The electric field between two parallel plates can be calculated using the formula E = V/d, where E is the electric field strength, V is the potential difference between the plates, and d is the distance between the plates.

3. What is the direction of the electric field between two parallel plates?

The direction of the electric field between two parallel plates is perpendicular to the surface of the plates, pointing from the positive plate to the negative plate.

4. How does the distance between the two parallel plates affect the electric field?

The electric field between two parallel plates decreases as the distance between the plates increases. This is because the potential difference (V) remains constant while the distance (d) increases, resulting in a smaller electric field strength (E).

5. Can the electric field between two parallel plates be uniform?

Yes, the electric field between two parallel plates can be uniform if the plates are of infinite size and the distance between them is constant. In this case, the electric field strength will be the same at all points between the plates.

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