A Physics False impression with Gauss’ Regulation

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 subject power 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 circumstances. On this Perception, we focus on the underlying assumptions required and what Gauss’ regulation really says.

What Gauss’ regulation says

Gauss’ regulation on integral kind relates the flux of the electrical subject 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.

You will need to 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 under.

The flux of the electrical subject ##vec E## via a small floor ingredient ##dS## is the amount spanned by ##vec E## and ##dS##. Solely the part 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 part 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 due to this fact 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 outcome nevertheless it doesn’t inform the entire story. The spherical symmetry implies that the electrical subject should be within the kind $$vec E = E(r) vec e_r.$$ 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 due to this fact categorical the electrical subject on the sphere as $$vec E = frac{Q_{rm enc}}{4pi varepsilon_0 R^2}.$$

Thus far, all the things appears to be like completely superb. Nevertheless, there are a couple of key factors to this which might be missed by the argument at the start 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 sphere power was the identical on our whole floor. It was essential to choose a floor that revered the symmetry.
• The symmetry additionally implied that the sphere was orthogonal to the floor in all places.

In a spherically symmetric setting, the electrical subject is all the time 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##. Nevertheless, every level was vital 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 under.

A floor (inexperienced) that doesn’t respect the spherical symmetry. The sphere 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 in all places.

The conclusion at the start 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 typically fail to recreate the argument within the case of cylinder symmetry. Particularly, that is executed for a line cost of linear cost density ##rho_ell##. The argument sometimes takes the next kind:

Because of the 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 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 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 count on the sphere power to rely on the size of the cylinder we selected. The dependence on ##R## is nevertheless completely superb 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 due to this fact ##Phi_{rm aspect} = E(R) 2pi RL##. On the top 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 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 right are:

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