A thin wire of infinite extent has a charge per unit length of λ. Using the cylindrical Gaussian

A thin wire of infinite extent has a charge per unit length...

A thin wire of infinite extent has a charge per unit length of $\lambda$ . Using the cylindrical Gaussian surface shown in FIGURE $19-46,$ show that the electric field produced by this wire at a radial distance $r$ has a magnitude given by $$E=\frac{\lambda}{2 \pi \epsilon_{0} r}$$ Note that the direction of the electric field is always radially away from the wire.

A thin wire of infinite extent has a charge per unit length of $\lambda$ .
Using the cylindrical Gaussian surface shown in FIGURE $19-46,$ show
that the electric field produced by this wire at a radial distance $r$
has a magnitude given by
$$E=\frac{\lambda}{2 \pi \epsilon_{0} r}$$
Note that the direction of the electric field is always radially away
from the wire.

Key Concepts

Charge per Unit Length

Charge per unit length, typically denoted by ?, characterizes the distribution of charge along a wire or line. It quantifies how much charge is present per meter (or any unit of length) of the wire, and it is a key parameter in determining the magnitude of the electric field produced by an infinite line charge.

Cylindrical Symmetry

Cylindrical symmetry in electrostatics means that the system's physical properties are invariant under rotations about and translations along a given axis. This symmetry implies that the electric field points radially outward and depends only on the distance from the axis of the wire, greatly simplifying the calculations using Gauss's Law.

Cylindrical Gaussian Surface

A cylindrical Gaussian surface is used when dealing with problems exhibiting cylindrical symmetry, such as an infinite line charge. By choosing a cylindrical surface whose axis coincides with the charged wire, the evaluation of the electric flux becomes straightforward because the electric field has a constant magnitude on the curved surface and is perpendicular to it.

Gauss's Law

Gauss's Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. This fundamental law of electromagnetism provides a powerful method for calculating electric fields in situations where symmetry allows for simplification of the surface integral.

Transcript

00:01 Question 68 states that a thin wire of infinite extent has a charge per unit length of lambda, using cylindrical gaussian surface shown in my diagram here, however, it may be better represented in figure 1946.

00:15 In your textbook, shown that the electric field produced by this wire at a radial distance r has a magnitude given by, i'll write this equation out, so the electric field, it's lambda, this linear charge density or charge per unit length, over 2 times pi, epsilon knot.

00:38 So this is what we want to solve for.

00:41 It does a little hint at the bottom.

00:43 It says, note that the direction of the electrical field is radially away from the wire itself.

00:50 So because it's positively charged, it goes in all directions away from the line.

00:59 So the blue here, blue cylinder, which i drew, is a gaussian surface around this positively charged.

01:08 Wire of charge.

01:10 So remember gauss's law, we state that the area times the allergic field is equivalent to the charge enclosed over epsilon not.

01:24 So this is an infinitely long wire.

01:29 So the amount of charge enclosed is not directly known because, i mean, you take a certain amount, a certain length of the wire, and a certain charge density of the wire.

01:39 So that means in terms of how much we have enclosed, while the actual amount in terms of coulums is not known, when i'm charged, we can say, well, if we know that it has a charge per unit length of lambda, as given in the question, if you multiply it by l, the length, so that's charge, say charging coulombs per length is what we have for lambda.

02:03 If we multiply that by the length, that would give us units of churons, which is of charge.

02:10 So that means if we take this lambda times l, that would be equivalent to the charge enclosed within our gaussian cylinder.

02:21 So on our right -hand side of our equation, we can represent q enclosed by this value lambda times l.

02:29 Obviously, epsilon not doesn't change.

02:32 We're solving for electric field, so that doesn't change either.

02:36 But we're going to represent the surface area of our cylinder.

02:42 So we know surface area of a circle, let's say, like the one surface of the surface of the one surface of the one surface of the cylinder.

02:47 Of our cylinder, well that's just pi r, but there's two of those.

02:56 So the surface area of both sides of our cylinder...

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