Question

A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$

Short answer

Short Answer: The electric field due to a thin sheet of uniform charge density can be calculated using Gauss's law. For a finite sheet, the field lines are perpendicular to the surface and evenly spaced, while for an infinite sheet, they are parallel, representing a uniform electric field. The electric field strength near a sheet of uniform charge density is given by \(E = \frac{\sigma}{2\epsilon_0}\), where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the vacuum permittivity. This electric field strength does not depend on the distance from the infinite sheet or on the distance near a finite sheet, when far from its edges.

Step-by-step solution

Sketch field lines due to the finite sheet of charge

To sketch the field lines due to the finite sheet of charge, imagine a small square in the sheet, where we will surround it with a Gaussian surface. The Gaussian surface should be a small cuboidal box extending on both sides of the sheet. Since the sheet has uniform charge density, the field lines will be equally spaced, and they will be perpendicular to the surface of the sheet. The field lines will point away from the positive charges and towards the negative charges.

Sketch field lines for the infinitely large sheet of charge

For the infinitely large sheet, the field lines will still be perpendicular to the sheet's surface. However, as the sheet becomes infinitely large, the field lines will be parallel to each other. These lines represent a uniform electric field.

Determine field strength dependence on the distance from the infinite sheet

In the case of the infinite sheet, the field strength does not depend on the distance from the sheet. The electric field is constant and will remain constant at any distance. This can be deduced from the parallel field lines in the sketch.

Approximating finite sheet as infinite sheet

For points close to the finite sheet and far from its edges, the field lines will appear nearly parallel. In this case, a finite sheet can be approximated as an infinitely large sheet. This means that for these points, the electric field strength will not depend on the distance from the sheet either.

Use Gauss's law to calculate the electric field near a sheet of uniform charge density

We will use Gauss's law to calculate the electric field: \(\oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_0}\), where \(Q_{enc}\) is the charge enclosed by the Gaussian surface, \(\vec{E}\) is the electric field, \(\vec{dA}\) is the differential area vector, and \(\epsilon_0\) is the vacuum permittivity. For a small cuboidal Gaussian surface enclosing a portion of the sheet, only the two faces parallel to the sheet contribute to the integral. The electric field is uniform and parallel to these faces, so the integral reduces to: \(E \cdot 2A = \frac{Q_{enc}}{\epsilon_0}\). The enclosed charge can be represented as \(Q_{enc} = \sigma A\), where \(\sigma\) is the surface charge density of the sheet and \(A\) is the area of each face of the Gaussian surface. Substituting this into the equation, we get: \(E \cdot 2A = \frac{\sigma A}{\epsilon_0}\). Solving for \(E\), we find the electric field to be: \(E = \frac{\sigma}{2\epsilon_0}\).