Will you be shocked if you're on the outside of a Faraday Cage and the electrical source is on the inside?

No, you are not safe. Shock city. The way a Faraday cage works is that the charges in the metal move to cancel out the electric field inside the cage; for instance, if a positive external charge is on the left of the cage, the free electrons in the metal will move to the left, and so any object inside the cage will experience a field directed to the right due to the original charge, and a field to the left due to the rearrangement of electrons. The fields cancel exactly (if the metal is a perfect conductor). On the other hand, if there is a positive charge inside the cage, the electrons in the metal will be attracted to the inner surface of the cage, leaving a net positive charge on the outside surface. This filed is in the same direction as that of the charge, because the outside, positive surface is slightly closer to any external object, and has the effect of amplifying the electric field on the outside of the cage. Obviously, this effect is usually negligible, but it shows that the cage certainly does not cancel out the charge.

A Faraday cage is designed to block electromagnetic radiation, not direct current. This means the thing that is being blocked is radiated through space, like a radio waves from a transmitter; the cage is not conducting electricity, like if the cage was plugged into a wall socket.

Whether the transmitter is on the outside and you are on the inside, or vice versa, the radiated energy cannot pass through the cage, so you are safe.

Jayne, above, is correct, with emphasis on “this effect is usually negligible,” meaning the power of the electromagnetic radiation is usually so low, currents in the cage itself are very tiny.

@PerryDolia; I’m confused, you say I’m correct, after disagreeing with me by saying it doesn’t matter which side of the cage you are on. By “negligible”, I was referring to the amplifying affect on a charge inside the cage; the electric field produced by the charge (or electromagnetic field if it is oscillating) will still be able to pass through the cage without interference if it is inside the cage, and if the charge is strong enough, it will shock you; you are only safe if you are the one in the cage.

Jayne,

Respectfully, that not how I understand it. A Faraday cage is grounded, it does not build hardly any current from the field it is exposed to, the charge is drained away to ground.

As I get it, if the transmitter is on the inside, the electromagnetic field is stopped by the cage. If the transmitter is on the outside, the electromagnetic field is stopped by the cage. The field does not pass through the cage, that is what the Faraday cage is for—to stop the field.

I agree that there is a small movement of electrons in the material of the cage, creating some (trivial) current.

I do not mean this as a debate, rather an explanation of how I understand it.

OK, a Faraday cage will protect against an internal charge, but only if it is grounded, which is not always the case.

Thank you Jane and PerryDolia for your answers!
It’s definitely cleared up my confusion on how a Faraday Cage works.