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In summary, the electric field between two charged plates is constant in the ideal case of infinite plates, but there may be edge effects for finite plates. As long as the width of the plates is much larger than the separation, the electric field can be assumed to be constant. This is because the potential varies linearly and the electric field is inversely proportional to distance. The behavior of a single charged plate can be understood by considering an infinite plane with a uniform charge density. The total flux passing through a cylinder placed on the plane does not depend on its length due to symmetry. The contributions from charges off to the side cancel out, resulting in a constant electric field. This can also be shown using cylindrical coordinates and an integral over the charge density

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I imagine if the point in consideration was very near one of the plates then the E-field would be stronger there as the electric field is inversely proportional to distance.

E = kQ/r.

For instance, "Is the electric field constant at all points on the line between two positionally fixed charges?" The equation E = kQ/r would suggest to me the answer is no.

What is going on here? Why is the E-field constant?

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The standard exposition is to consider an infinite plane with a uniform charge density. You can then use a cylinder as a gaussian surface and argue that by symmetry all the flux must pass through the ends. Note that the total flux does not depend on how long the cylinder is.

But if it helps to think in terms of component effects consider this. Imagine you are a test charge suspended over a flat level (infinite) sheet with a uniform charge distribution.

Yes the close you are to the plane the more the charge directly below you contributes to the E field. However since the charges off to the side are pushing at smaller opposite angles the more their contributions cancel out. Only their components perpendicular to the plane contribute. This decrease offsets the other increase exactly.

Ultimately remember there is a charge density and not a finite point charge at each point. The closer you get the less charge is within a thin cone below you. As you almost touch the charge within a cone below you approaches zero.

It is a tricky integral but you can (I think) add up the contributions for each infinitesimal area of the plane and see that it is independent of the distance.

Hmmm... let's see using cylindrical coordinates with surface charge density [itex]\rho[/itex] on the plane [itex] z = 0[/itex]

[tex] (z,r,\theta)[/tex]

figure the electrostatic potential at a point on the z-axis.

The charge of a particular differential area is [itex]dQ = \rho r\cdot dr \cdot d\theta[/itex].

The distance to the point on the z-axis is [itex]R=\sqrt(r^2 + z^2)[/itex]

The contributed electrostatic force is:

[tex] |dE| = \frac{kdQ}{R^2}[/tex]

But the only contribution in the vertical direction will be the z-component which is

[tex] dE_z = \frac{kdQ}{R^2}\cos(\phi)[/tex]

where [itex]\phi[/itex] is the angle the hypotenuse makes with the z-axis.

This angle will be

[tex] \cos(\phi)=\frac{z}{R}[/tex].

Now we are ready to set up the integral:

[tex]E_z = \int dQ\frac{kz}{R^3} = k \rho \int \frac{z r dr d\theta}{(z^2 + r^2)^{3/2}}[/tex]

[tex] = 2\pi k\rho \int_{r=0}^{\infty} \frac{z r dr}{(z^2 + r^2)^{3/2}}[/tex]

Now use the trig substitution [itex] r = z\tan(\alpha)[/itex] and [itex]dr = z\sec^2(\alpha)d\alpha[/itex]

You get:

[tex]E_z=2\pi k\rho \int_{\alpha=0}^{\pi/2} \frac{z^3 \tan(\alpha)\sec^2(\alpha) d\alpha}{z^3\sec^3(\alpha)}[/tex]

[tex]E_z=2\pi k\rho \int_{\alpha=0}^{\pi/2}\sin(\alpha)d\alpha[/tex]

[tex]E_z=2\pi k\rho[\cos(0)-\cos(\pi/2)]=2\pi k\rho[/tex]

The z dependence exactly cancels. This because you get that extra z/R contribution from the angle off the z-axis. In other word the higher up you are the more charge is almost below you instead of way off to the sides and canceling out. Yet the farther you are from it so the E field gets neither stronger nor weaker.

Hope that helps.

But if it helps to think in terms of component effects consider this. Imagine you are a test charge suspended over a flat level (infinite) sheet with a uniform charge distribution.

Yes the close you are to the plane the more the charge directly below you contributes to the E field. However since the charges off to the side are pushing at smaller opposite angles the more their contributions cancel out. Only their components perpendicular to the plane contribute. This decrease offsets the other increase exactly.

Ultimately remember there is a charge density and not a finite point charge at each point. The closer you get the less charge is within a thin cone below you. As you almost touch the charge within a cone below you approaches zero.

It is a tricky integral but you can (I think) add up the contributions for each infinitesimal area of the plane and see that it is independent of the distance.

Hmmm... let's see using cylindrical coordinates with surface charge density [itex]\rho[/itex] on the plane [itex] z = 0[/itex]

[tex] (z,r,\theta)[/tex]

figure the electrostatic potential at a point on the z-axis.

The charge of a particular differential area is [itex]dQ = \rho r\cdot dr \cdot d\theta[/itex].

The distance to the point on the z-axis is [itex]R=\sqrt(r^2 + z^2)[/itex]

The contributed electrostatic force is:

[tex] |dE| = \frac{kdQ}{R^2}[/tex]

But the only contribution in the vertical direction will be the z-component which is

[tex] dE_z = \frac{kdQ}{R^2}\cos(\phi)[/tex]

where [itex]\phi[/itex] is the angle the hypotenuse makes with the z-axis.

This angle will be

[tex] \cos(\phi)=\frac{z}{R}[/tex].

Now we are ready to set up the integral:

[tex]E_z = \int dQ\frac{kz}{R^3} = k \rho \int \frac{z r dr d\theta}{(z^2 + r^2)^{3/2}}[/tex]

[tex] = 2\pi k\rho \int_{r=0}^{\infty} \frac{z r dr}{(z^2 + r^2)^{3/2}}[/tex]

Now use the trig substitution [itex] r = z\tan(\alpha)[/itex] and [itex]dr = z\sec^2(\alpha)d\alpha[/itex]

You get:

[tex]E_z=2\pi k\rho \int_{\alpha=0}^{\pi/2} \frac{z^3 \tan(\alpha)\sec^2(\alpha) d\alpha}{z^3\sec^3(\alpha)}[/tex]

[tex]E_z=2\pi k\rho \int_{\alpha=0}^{\pi/2}\sin(\alpha)d\alpha[/tex]

[tex]E_z=2\pi k\rho[\cos(0)-\cos(\pi/2)]=2\pi k\rho[/tex]

The z dependence exactly cancels. This because you get that extra z/R contribution from the angle off the z-axis. In other word the higher up you are the more charge is almost below you instead of way off to the sides and canceling out. Yet the farther you are from it so the E field gets neither stronger nor weaker.

Hope that helps.

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An electric field is a physical quantity that describes the influence that a charged object has on other charged objects. It is a vector quantity, meaning it has both magnitude and direction.

The electric field between two charged plates can be calculated using the equation 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.

In ideal conditions, the electric field between two charged plates is constant. However, in real-world situations, factors such as the size and shape of the plates, as well as external influences, can cause the electric field to vary slightly.

A constant electric field between two charged plates allows for a uniform force to be exerted on charged particles between the plates. This is important in many applications, such as in capacitors and particle accelerators.

Yes, if the potential difference between the plates is zero, then the electric field between them will also be zero. This can happen when the plates are connected to a conductor that is at the same potential as the plates, or if the plates are not charged at all.

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