Electric field inside a cavity (Faraday Cage)

Question

David J. Griffiths in his "Introduction to Electrodynamics" (chapter2), says that the Electric field inside a cavity (in a conductor) which contains a point charge is nonzero. this is because we can choose a gaussian surface inside the cavity (just a little smaller than the cavity), then by applying Gauss's law:
$$\int \vec E \, .d\vec a = \frac{q}\epsilon_0 \rightarrow \vec E\neq0$$

and when the charge is outside of the conductor (the case of Faraday Cage), he argues that if we choose a closed path which is partly in the "meat" of the conductor and partly in the cavity, then we can prove that the Field is zero in the latter reigion ($C_1$ is the path in the conductor and $C_2$ is the path in the cavity):

$$\vec \nabla \times \vec E=0 \Rightarrow \int_{C}\vec E. \vec dl=\int_{C_1} \vec E_1.\vec dl +\int_{C_2}E_2.\vec dl=0$$

since $E_1$ (the field inside the conductor) is zero, $E_2$ should be zero as well.

can someone plaese explain his second argument? I don't really understand it, what exactly is preventing us to apply this argument to the first case? conceptually and mathematically (in the context of electrostatics)

Thanks in advance.

Answer 1

$\int_{C}\vec E. \vec dl$ represent the potential drop a charge would experience when it moves on a path defined as $C$. In this case, the path $C$ is indeed a closed loop. That is, where you start and where you stop are the same positions. Hence, the potential drop is zero.

Thus, $$\int_{C}\vec E. \vec dl=0 $$

Now, on dividing the path $C$ in $C_1$ and $C_2$, if electric field is zero in path $C_2$, moving on it, a charge can never experience a potential drop. So, both end points of path $C_2$ are at same potential, and so are the end points of point $C_1$. Hence potential drop about $C_1$ is $0$

As potential drop about path $C$ is given as $\int_{C}\vec E. \vec dl$, potential drop about $C_1$ will be,

$$\int_{C_1}\vec E. \vec dl=0$$

Now, you could obtain this integral as $0$ in three ways

• $\vec E$ is perpendicular to $\vec dl$ at each and every point on path $C_1$. This would be irrational to argue as you could have chosen the path in arbitrarily any direction at a point and it can't be possible that $\vec E$ is perpendicular to the path at every case. So, this possibility is out of the zone of consideration

• End points of path $C_1$ are same and it forms a closed loop, which is possible not going to be the case. So, let's move on.

• The $\vec E$ becomes $0$ instead.

Oh no! Now we're just left with only possibility left so it's probability will be $100\%$.

Note: you might argue that $\vec E$ is variable inside the conductor and is so oriented that every path forms a net zero potential drop. Well, a field inside the conductor would generate a current and that would no longer be the case of electrostatics.