## 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 energy on the floor is subsequently ##E = Q/varepsilon_0 A##.

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

## What Gauss’ regulation says

Gauss’ regulation on integral type relates the flux of the electrical area via 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.

It is very important notice that the flux integral solely relies on 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 beneath.

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 realm of ##S##. The argument within the introduction is subsequently reliant on these two assumptions.

## Spherical symmetry

Does the argument work within the case of spherically symmetric cost distribution? Nicely … Sure and no. The argument does give the proper end result however it doesn’t inform the entire story. The spherical symmetry implies that the electrical area should 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. Because the space of the sphere is ##A = 4pi R^2##, it follows that ##Phi = E(R) A = 4pi E(R) R^2##. We will subsequently specific the electrical area on the sphere as $$vec E = frac{Q_{rm enc}}{4pi varepsilon_0 R^2}.$$

Thus far, every part seems to be completely effective. Nevertheless, there are just a few key factors to this which are missed by the argument initially of the Perception:

- We wanted to make use of a symmetry to conclude the useful type of the electrical area.
- From the symmetry, we might additionally conclude that the sphere energy was the identical on our whole floor. It was essential to select a floor that revered the symmetry.
- The symmetry
*additionally*implied that the sphere 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’d have chosen some other floor, see the determine beneath.

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

## Cylinder symmetry

College students usually fail to recreate the argument within the case of cylinder symmetry. Particularly, that is achieved 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 beneath.*

*The realm of the cylinder is the sum of the aspect 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 subsequently $$E = frac{rho_ell L}{2pi R(L+R)}.$$*

This end result inherently smells a bit dangerous. We must always not anticipate the sphere energy to rely on the size of the cylinder we selected. The dependence on ##R## is nevertheless completely effective 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 aspect of the cylinder, the sphere can certainly be argued to be each orthogonal to the floor and of fixed magnitude. The flux via the aspect is subsequently ##Phi_{rm aspect} = E(R) 2pi RL##. On the tip caps, the sphere is parallel to the floor so the flux is ##Phi_{rm caps} = 0##. Consequently, the whole flux is $$Phi = E(R) 2pi RL$$. Because the enclosed cost is ##Q_{rm enc} = rho_ell L##, this suggests $$E(R) = frac{rho_ell L}{2pi varepsilon_0 RL} = frac{rho_ell}{2pi varepsilon_0 R}.$$

## Watch out!

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

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

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