A single isolated, large conducting plate (Fig. 22.39)...

A single isolated, large conducting plate (Fig. 22.39) Figure 22.39 Problem 22.51. has a charge per unit area $\sigma$ on its surface. Because the plate is a conductor, the electric field at its surface is perpendicular to the surface and has magnitude $E=\sigma / \epsilon_0$ (a) In Example 22.7 (Section 22.4) it was shown that the field caused by a large, uniformly charged sheet with charge per unit area $\sigma$ has magnitude $E=\sigma / 2 \epsilon_{00}$ exactly half as much as for a charged conducting plate. Why is there a difference? (b) Regarding the charge distribution on the conducting plate as being two sheets of charge (one on each surface), each with charge per unit area $\sigma$, use the result of Example 22.7 and the principle of superposition to show that $E=0$ inside the plate and $E=\sigma / \epsilon_0$ outside the plate. (FIGURE CANT COPY)

Key Concepts

Superposition Principle

The superposition principle in electromagnetism holds that the net electric field due to multiple charge distributions is the vector sum of the fields produced by each distribution individually. This concept allows one to analyze complex problems by breaking them into simpler parts, such as summing the contributions from two oppositely facing charged sheets.

Conductors and Electric Fields

In electrostatics, conductors achieve equilibrium by redistributing charges so that the electric field inside the material is zero. Charges reside on the surface, and the resulting field just outside the conductor is perpendicular to the surface. This is a fundamental property of conductors that distinguishes them from other types of charge distributions.

Electric Field of a Charged Sheet

A single, infinitely large sheet of charge produces a uniform electric field of magnitude ?/(2??) on each side. This result arises from Gauss’s law and assumes an idealized sheet of charge, where the effects of edge fields are neglected.

Charge Distribution on Conductors and Field Cancellation

In a conducting plate, charges reside on both surfaces; however, the fields produced by these surface charges cancel inside the conductor, resulting in zero electric field. Viewing the conductor as two separate charged sheets leads to the conclusion that the net external field is the sum of the fields from each sheet, thereby doubling the field magnitude compared to a single sheet.

Transcript

00:01 So we have two metallic plates and we have some plus positive charge and negative charge on them.

00:13 Let me draw this properly.

00:17 Yeah, that is good enough.

00:18 Now, because there is positive charge on this surface and negative charge of this surface, they attract each other, which means all the positive charge on the left plate will come to the surface to its right surface to get as close as possible to the negative charges.

00:36 And all the negative charges on the right plate comes to its left surface to get as close as possible to the positive charges.

00:46 But because the plates are made of conductors, there can be no charge inside, which means all of the charge will reside within these inside surfaces.

00:58 Now for part a, we want to find out the field within this region.

01:02 Now because the distance between the plates is much small, very small compared to the diameters, of the plates themselves, we can think of these as infinite surfaces.

01:13 Now because this is an infinite surface and there is translational symmetry, the field must be in this direction.

01:24 That is, let's say if so, now let us consider a gaussian surface that looks like a cylinder which goes like this, like this.

01:41 Now the total flux will be integral e.

01:48 D a which is equal to charge enclosed divided by epsilon not now let's divide this up into three surfaces surface number one is inside the middle the circular surface the second circular surface is outside the middle the third surface is the body of the sphere now on the first surface the the electric field is zero to find first let's write this out the flux will be is integrated eda, which means the total flux is the sum of flux on first surface plus the flux on second surface, on third surface.

02:40 Now on surface number one, the electric field is simply zero because it's inside the metal, which means the flux must be zero.

02:48 Whereas on surface number three, the normal of the area will always be perpendicular to the electric field.

02:55 So the field, the flux there is also zero, which means the only flux will be due to surface 2.

03:04 Now because the area vector of surface 2 is in this direction and the electric field is also in this direction, the angle between them is 0 and we can write the flux as simply integral eda, which is e times the area of the surface number 2.

03:21 We know that this is equal to qn close divided by epsilon not.

03:24 Hence, the electric field magnitude would be q enclosed divided by a by epsilon or not...

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