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Chapter 21: Problem 11

Why can't you use Gauss's law to determine the field of a uniformly charged cube? Why couldn't you use a cubical Gaussian surface?

Short Answer

Expert verified
Gauss's Law cannot be used to determine the field of a uniformly charged cube and a cube cannot be used as a Gaussian surface because a cube does not provide the symmetry necessary for Gauss's law to be applicable, which is that at every point on the Gaussian surface, the electric field magnitude and the angle between the electric field vector and the area vector should be same.

Step by step solution

01

Understanding Gauss's Law

The first step is to understand what Gauss's law is. Gauss's Law links the electric field surrounding a charge to the total charge. It states that the total electric flux out of a closed surface is equal to the charge enclosed by that surface divided by the permittivity.
02

Understanding Symmetry in Gauss's Law

For Gauss鈥檚 Law to be applicable, the symmetry of the system should be such that every point on the Gaussian surface has the same electric field magnitude and the same angle between the electric field vector and the area vector. This is possible in the case of spherically, cylindrically or planar symmetric charge distributions.
03

Evaluating the Problem

In our case, a cube does not provide such symmetry. For instance, if you place a cubical Gaussian surface inside a uniformly charged cube, the electric field at the center towards the edges of the cube is less than the electric field at the center towards the corners. Hence, the symmetry required for Gauss's law is absent, and we can't use a cubical Gaussian surface.

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Most popular questions from this chapter

A solid sphere \(25 \mathrm{cm}\) in radius carries \(14 \mu \mathrm{C},\) distributed uniformly throughout its volume. Find the electric field strength (a) \(15 \mathrm{cm},\) (b) \(25 \mathrm{cm},\) and (c) \(50 \mathrm{cm}\) from its center.

Why must the electric field be zero inside a conductor in electrostatic equilibrium?

The electric flux through a closed surface is zero. Must the electric field be zero on that surface? If not, give an example.

An infinitely long solid cylinder of radius \(R\) carries a nonuniform charge density given by \(\rho=\rho_{0}(r / R),\) where \(\rho_{0}\) is a constant and \(r\) is the distance from the cylinder's axis. Find an expression for the magnitude of the electric field as a function of position \(r\) within the cylinder.

An electron close to a large, flat sheet of charge is repelled from the sheet with a 1.8-pN force. Find the surface charge density on the sheet.
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