## Introduction

It’s comparatively frequent to see the next kind of argument:

The floor space is ##A## and the enclosed cost is ##Q##. The electrical area power on the floor is due to this fact ##E = Q/varepsilon_0 A##.

The issue is that this assertion is just true in very particular instances. On this Perception, we talk about the underlying assumptions required and what Gauss’ legislation truly says.

## What Gauss’ legislation says

Gauss’ legislation on integral type relates the flux of the electrical area by a closed floor to the cost enclosed by the floor ##varepsilon_0 Phi = Q_{rm enc}##. Right here ##Phi## is the flux, ##Q_{rm enc}## the enclosed cost, and ##varepsilon_0## the permittivity of vacuum. In barely extra mathematical phrases, $$varepsilon_0 oint_S vec E cdot dvec S = int_V rho , dV,$$ the place ##S## is the floor, ##V## the enclosed quantity, and ##rho## the cost density.

You will need to word that the flux integral solely is determined by the part of ##vec E## orthogonal to the floor. Any part parallel to the floor won’t result in a flux out of the floor, see the determine under.

Mathematically, if ##vec E## is orthogonal to and of fixed magnitude ##E_0## on the floor, then ##vec E cdot dvec S = E_0 dS##. In that case $$Phi = oint_S E_0 dS = E_0 oint_S dS = E_0 A,$$ the place ##A## is the world of ##S##. The argument within the introduction is due to this fact reliant on these two assumptions.

## Spherical symmetry

Does the argument work within the case of spherically symmetric cost distribution? Effectively … Sure and no. The argument does give the right end result however it doesn’t inform the entire story. The spherical symmetry implies that the electrical area have to be within the type $$vec E = E(r) vec e_r.$$ Utilizing a sphere of radius ##R## as our floor ##S##, ##vec E## is orthogonal to the floor in all places. For the reason that space of the sphere is ##A = 4pi R^2##, it follows that ##Phi = E(R) A = 4pi E(R) R^2##. We are able to due to this fact specific the electrical area on the sphere as $$vec E = frac{Q_{rm enc}}{4pi varepsilon_0 R^2}.$$

To date, all the things appears completely positive. Nevertheless, there are just a few key factors to this which can be missed by the argument originally of the Perception:

- We would have liked to make use of a symmetry to conclude the practical type of the electrical area.
- From the symmetry, we might additionally conclude that the sector power was the identical on our whole floor. It was essential to select a floor that revered the symmetry.
- The symmetry
*additionally*implied that the sector was orthogonal to the floor in all places.

Due to these three factors, we might certainly conclude that ##E = Q/varepsilon_0 A##. Nevertheless, every level was essential to attract this conclusion.

## Failure of the argument for spherical symmetry

The conclusion ##E = Q/varepsilon_0 A## would fail for spherical symmetry if we might have chosen some other floor, see the determine under.

For instance, if we had taken a dice as an alternative of a sphere, the electrical area would:

- Not have fixed magnitude on the floor.
- Not be orthogonal to the floor in all places.

The conclusion originally of the Perception would conclude that ##E = Q_{rm enc}/6varepsilon_0 L^2##, the place ##L## is the dice’s facet size, which might be incorrect.

## Cylinder symmetry

College students typically fail to recreate the argument within the case of cylinder symmetry. Particularly, that is carried out for a line cost of linear cost density ##rho_ell##. The argument usually takes the next type:

*Because of the cylinder symmetry, select a cylinder floor of radius ##R## and size ##L##, see the determine under.*

*The realm of the cylinder is the sum of the facet space and the bottom areas ##A = 2pi RL + 2pi R^2 = 2pi R(L+R)##. The cost enclosed by the floor is ##Q_{rm enc} = rho_ell L##. The electrical area is due to this fact $$E = frac{rho_ell L}{2pi R(L+R)}.$$*

This end result inherently smells a bit dangerous. We should always not anticipate the sector power to depend upon the size of the cylinder we selected. The dependence on ##R## is nonetheless completely positive as symmetry arguments would indicate that ##vec E = E(rho) vec e_rho##. Right here ##rho## is the radial coordinate in cylinder coordinates and ##vec e_rho## is the corresponding foundation vector.

So the place does the argument fail? On the facet of the cylinder, the sector can certainly be argued to be each orthogonal to the floor and of fixed magnitude. The flux by the facet is due to this fact ##Phi_{rm facet} = E(R) 2pi RL##. On the top caps, the sector is parallel to the floor so the flux is ##Phi_{rm caps} = 0##. Consequently, the entire flux is $$Phi = E(R) 2pi RL$$. For the reason that enclosed cost is ##Q_{rm enc} = rho_ell L##, this means $$E(R) = frac{rho_ell L}{2pi varepsilon_0 RL} = frac{rho_ell}{2pi varepsilon_0 R}.$$

## Watch out!

In conclusion, be very cautious when making use of ##E = Q_{rm enc}/varepsilon_0 A##. The necessities for this being appropriate are:

- The realm needs to be that of a floor on which the electrical area is orthogonal to the floor.
- The electrical area ought to have a relentless magnitude in that space.
- The flux by some other surfaces concerned in making a closed floor needs to be zero.

Professor in theoretical astroparticle physics. He did his thesis on phenomenological neutrino physics and is at present additionally working with completely different elements of darkish matter in addition to physics past the Customary Mannequin. Writer of “Mathematical Strategies for Physics and Engineering” (see Perception “The Beginning of a Textbook”). A member at Physics Boards since 2014.