Conductors

(617) |

inside a conductor. It immediately follows from the Maxwell equation, , that

(619) |

However, if and lie inside the same conductor then it is clear from Eq. (620) that the potential difference between and is zero. This is true no matter where and are situated inside the conductor, so we conclude that the scalar potential must be

Not only is the electric field inside a conductor zero. It is also possible to demonstrate that the field within an empty cavity lying inside a conductor is also zero, provided that there are no charges within the cavity. Let us, first of all, apply Gauss' law to a surface which surrounds the cavity, but lies wholly in the conducting material (see Fig. 41). Since the electric field is zero in a conductor, it follows that zero net charge is enclosed by . This does not preclude the possibility that there are equal amounts of positive and negative charges distributed on the inner surface of the conductor. However, we can easily rule out this possibility using the steady-state relation

for any closed loop . If there are any electric field-lines inside the cavity then they must run from the positive to the negative surface charges. Consider a loop which straddles the cavity and the conductor, such as the one shown in Fig. 41. In the presence of field-lines, it is clear that the line integral of along that portion of the loop which lies inside the cavity is non-zero. However, the line integral of along that portion of the loop which runs through the conducting material is obviously zero (since inside a conductor). Thus, the line integral of the field around the closed loop is non-zero. This, clearly contradicts Eq. (621). In fact, this equation implies that the line integral of the electric field along any path which runs through the cavity, from one point on the interior surface of the conductor to another, is zero. This can only be the case if the electric field itself is zero everywhere inside the cavity. There is one proviso to this argument. The electric field inside a cavity is only zero if the cavity contains no charges. If the cavity contains charges then our argument fails because it is possible to envisage that the line integral of the electric field along many different paths across the cavity could be zero without the fields along these paths necessarily being zero (this argument is somewhat inexact: we shall improve it later on).

We have shown that if a cavity is completely enclosed by a conductor then no stationary distribution of charges outside can ever produce any fields inside. So, we can shield a piece of electrical equipment from stray external electric fields by placing it inside a metal can. Using similar arguments to those given above, we can also show that no static distribution of charges inside a closed conductor can ever produce a field outside the conductor. Clearly, shielding works both ways!

Let us consider some small region on the surface of a conductor. Suppose that the local surface charge density is , and that the electric field just outside the conductor is . Note that this field must be directed

(622) |

as the relationship between the normal electric field immediately outside a conductor and the surface charge density.

Let us look at the electric field generated by a sheet charge distribution a little more carefully. Suppose that the charge per unit area is . By symmetry, we expect the field generated below the sheet to be the mirror image of that above the sheet (at least, locally). Thus, if we integrate Gauss' law over a pill-box of cross sectional area , as shown in Fig. 43, then the two ends both contribute to the surface integral, where is the normal electric field generated above and below the sheet. The charge enclosed by the pill-box is just . Thus, Gauss' law yields a symmetric electric field

So, how do we get the asymmetric electric field of a conducting surface, which is zero immediately below the surface (

(626) |

(627) | |||

(628) |

which is in agreement with Eq. (623).

The external field exerts a force on the charge sheet. The field generated locally
by
the sheet itself obviously cannot exert a force (the sheet cannot exert
a force on itself!). The force per unit area acting on the surface of the
conductor always acts outward, and is given by

(629) |

(630) |

(631) |

We have seen that
an electric field is excluded from the inside of the conductor, but not
from the outside, giving rise to a net *outward*
force. We can account for this by saying that the field exerts a
*negative* pressure
on the conductor.
We know that if we evacuate a metal can then the
pressure difference between the inside and the outside eventually causes
it to *implode*. Likewise, if we place the can in a strong electric field then
the pressure difference between the inside and the outside will eventually cause
it to *explode*.
How big a field do we need before the electrostatic pressure difference
is the same as that obtained by
evacuating the can? In other words, what field exerts a negative
pressure of one atmosphere (*i.e.*, newtons per meter squared) on conductors?
The answer is a field of strength volts per meter.
Fortunately, this is a rather
large field, so there is no danger of your car exploding when you turn on the
stereo!