A Physics False impression with Gauss’ Legislation

Introduction

It’s comparatively widespread to see the next sort of argument:

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

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

What Gauss’ regulation says

Gauss’ regulation on integral type relates the flux of the electrical subject by way of 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 observe 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 beneath.

The flux of the electrical subject ##vec E## by way of a small floor ingredient ##dS## is the amount 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 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? Properly … Sure and no. The argument does give the right outcome but it surely doesn’t inform the entire story. The spherical symmetry implies that the electrical subject 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 all over the place. 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 will due to this fact categorical the electrical subject on the sphere as $$vec E = frac{Q_{rm enc}}{4pi varepsilon_0 R^2}.$$

To date, every thing appears completely nice. Nonetheless, there are a couple of key factors to this which might be missed by the argument initially of the Perception:

• We wanted to make use of a symmetry to conclude the useful type of the electrical subject.
• 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 all over the place.

In a spherically symmetric setting, the electrical subject is at all times perpendicular to a spherical floor respecting the symmetry. It’s also 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 crucial 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 every other floor, see the determine beneath.

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 subject would:

• Not have fixed magnitude on the floor.
• Not be orthogonal to the floor all over the place.

The conclusion initially 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. Specifically, that is accomplished for a line cost of linear cost density ##rho_ell##. The argument sometimes takes the next type:

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

A cylinder (inexperienced) floor surrounding an infinite line cost (brown).

The world 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 subject is due to this fact $$E = frac{rho_ell L}{2pi R(L+R)}.$$

This outcome 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 facet of the cylinder, the sector can certainly be argued to be each orthogonal to the floor and of fixed magnitude. The flux by way of 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 whole flux is $$Phi = E(R) 2pi RL$$. For the reason that 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 appropriate are:

• The world ought to be that of a floor on which the electrical subject is orthogonal to the floor.
• The electrical subject ought to have a relentless magnitude in that space.
• The flux by way of every other surfaces concerned in making a closed floor ought to be zero.