Faraday cages

Question 1

how does a faraday cage work

Answer 1

the free charges inside the cage rearrange themselves to cancel out any external or internal electric field. there position makes it so the electric field inside the conducting material is 0

Question 2

prove that the induced charge at the inner wall of a cavity is equal and opposite to the charge inside the cavity

Answer 2

if the charge is +q in the cavity:

E inside the conductor is 0 (by definition)
hence ∮E⋅dA =0 (Gaussian surface integral for a surface containing the cavity)
meaning the flux is 0.

therefore the charge enclosed by the surface must be 0 so the induced charge at the cavity’s surface is -q

Question 3

if a conductor has 2 cavities and one contains a charge, what is the charge induced on the inner surface of the empty cavity

Answer 3

0. no charge is induced as the empty cavity is screened from the other cavity.

Question 4

How does the Biot-Savart law help to solve for the magnetic field of a wire with a cavity

Answer 4

take a cross-section of the wire (circle with a circular cavity.
The magnetic field at a point P equals the magnetic field created by the wire if it didn’t have the cavity, minus the magnetic field that would be created by a wire the size of the cavity

easiest as: B = Iµ₀/2πr
I: the current (can by swapped for j when taking the wire cross-section)
r: distance from the centre of the wire

Question 5

explain why a B-field is just a relativistic correction for an E-field

Answer 5

for a neutral wire, a moving test charge outside of it experiences no
E-field. When transforming into the charge’s frame of reference, the ions and electrons in the wire length contract by a different factor. Hence the charge per unit length is >0 (ions contracted more than electrons). This creates an E-field that acts on the test charge. in the lab frame, this is interpreted as a force due to the B-field

Question 6

prove an external charge doesn’t induce a charge on the walls of a cavity in a conductor

Answer 6

proof by falsification (assume it can happen):

consider a positive charge outside the conductor.
suppose a charge is induced on the cavity, negative closest to the external charge (point B) and positive on the opposite side (point A).

the total charge on the cavity surface is 0 by Gauss’ law (no E-field inside the conductor).
The induced charges will result in an E-field from A to B and will loop back to A through the conductor.
since E inside a conductor is 0 then ∫E dl (B –> A) = 0

The circulation of an E-field is 0:
∮E dl = ∫E dl (B –> A) + ∫E dl (A –> B) = 0
hence ∫E dl (A –> B) = 0
and so there must be no charges on the cavity surface

Question 7

how to draw the direction of a B-field relative the the page

Answer 7

circle with a X is INTO the page

circle with a dot is OUT of the page