HomePhysicsA Physics False impression with Gauss' Regulation

# A Physics False impression with Gauss’ Regulation

## Introduction

It’s comparatively frequent to see the next sort 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 barely true in very particular circumstances. 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 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 element of ##vec E## orthogonal to the floor. Any element parallel to the floor is not going to result in a flux out of the floor, see the determine under. The flux of the electrical area ##vec E## by a small floor aspect ##dS## is the quantity spanned by ##vec E## and ##dS##. Solely the element of ##vec E## perpendicular to the floor (in pink) is related as the grey area has the identical quantity as that delimited by the purple dashes. Any element parallel to ##dS## doesn’t contribute to the flux.

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? Properly … Sure and no. The argument does give the right end result nevertheless 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}.\$\$

To this point, all the pieces seems to be completely nice. Nonetheless, there are a couple of key factors to this which might be missed by the argument firstly of the Perception:

• We wanted to make use of a symmetry to conclude the useful type of the electrical area.
• From the symmetry, we may additionally conclude that the sector energy was the identical on our complete 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. In a spherically symmetric setting, the electrical area is at all times perpendicular to a spherical floor respecting the symmetry. Additionally it is of fixed magnitude on the floor.

Due to these three factors, we may certainly conclude that ##E = Q/varepsilon_0 A##. Nonetheless, 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 under. A floor (inexperienced) that doesn’t respect the spherical symmetry. The sector is neither of fixed magnitude nor perpendicular to the floor.

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 firstly 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 carried out for a line cost of linear cost density ##rho_ell##. The argument usually takes the next type:

As a result of cylinder symmetry, select a cylinder floor of radius ##R## and size ##L##, see the determine under. A cylinder (inexperienced) floor surrounding an infinite line cost (brown).

The world 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 should always not anticipate the sector energy to depend upon the size of the cylinder we selected. The dependence on ##R## is nonetheless completely nice as symmetry arguments would suggest 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 sector can certainly be argued to be each orthogonal to the floor and of fixed magnitude. The flux by the aspect is subsequently ##Phi_{rm aspect} = 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\$\$. Because the 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 world needs to 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 by some other surfaces concerned in making a closed floor needs to be zero.
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