# Electric field inside a cavity (Faraday Cage)

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)

$$\int_{C}\vec E. \vec dl$$ 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.

• Thanks. But sorry I should've explain my problem better. I can understand the second argument for the Faraday Cage, but what is preventing me to use this line integral for the first case when there's a charge inside the cavity? I also understand that from Laplace's eq. + uniqueness theorem that the constant potential is the answer and for the first case from Poisson's eq. that the constant potential isn't the answer for sure, so I just want to know that why line integral can't be applied to the first case. – Amtgh Feb 28 at 2:34