Q about electric field between two parallel plates

[asdff529]

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!

 

[ShayanJ]

Consider Gauss's law $\oint \vec{E}\cdot\hat{n}da=\frac{q}{\varepsilon_0}$.
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 $q=lw\sigma$ 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 $E lw$ and so we have $E=\frac{\sigma}{\varepsilon_0}$.
Now consider $V=-\int_a^b \vec{E}\cdot\vec{dr}$ 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 $E=\frac V d$. As you saw in the last paragraph,the electric field between the planes was constant along their separation and so the formula $E=\frac V d$ can be used in that case.For example you can have $V=\frac{d\sigma}{\varepsilon_0}$ 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.

 

[asdff529]

Consider Gauss's law $\oint \vec{E}\cdot\hat{n}da=\frac{q}{\varepsilon_0}$.
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 $q=lw\sigma$ 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 $E lw$ and so we have $E=\frac{\sigma}{\varepsilon_0}$.
Now consider $V=-\int_a^b \vec{E}\cdot\vec{dr}$ 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 $E=Vd$. As you saw in the last paragraph,the electric field between the planes was constant along their separation and so the formula $E=Vd$ can be used in that case.For example you can have $V=\frac{\sigma}{d\varepsilon_0}$ 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?

 

[ShayanJ]

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.

 

[sophiecentaur]

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.