Gauss' law

Question 1

equation of Gauss’ law and in words

Answer 1

∮ E⋅dA = Q(enc)/ε₀
E and dA are vectors (underlined)

E-field flux through a surface equals the charge enclosed divided by ε₀

Question 2

use Gauss’ law on a point charge to calculate its electrical field

Answer 2

The Gaussian surface is a sphere with a point charge at the centre

∮ E⋅dA = Q/ε₀

since E//dA:
∮ EdA = Q/ε₀

since E is constant across the surface
E∮dA = Q/ε₀
E4πr² = Q/ε₀
∴E= Q/4πr²ε₀

Question 3

use Gauss’ law on an infinite line that’s charged to find its electrical field

Answer 3

the gaussian surface is a cylinder with the line at the centre

∮ E⋅dA = Q/ε₀
= ∮ E⋅dA (top) + ∮ E⋅dA (Bottom) + ∮ E⋅dA (sides)
top and bottom refer to the circle faces and sides is the curved face

for the top and bottom E and dA are perpendicular so E⋅dA=0
for the sides E//dA so E⋅dA= EdA

E∮dA (sides)= Q/ε₀
∮dA (sides) =2πrl (l: the length of wire in the surface)
E2πrl= λl/ε₀ (Q=λl where λ is the charge density per unit length)
∴E= λ/2πrε₀

Question 4

use Gauss’ law on an infinite flat surface that’s charged to find its electrical field

Answer 4

the gaussian surface is a cylinder passing through the flat surface. the flat surface passes through halfway along the cylinder.

∮ E⋅dA = Q/ε₀
= ∮ E⋅dA (top) + ∮ E⋅dA (Bottom) + ∮ E⋅dA (sides)

for the top and bottom E//dA so E⋅dA= EdA
for the sides E and dA are perpendicular so E⋅dA=0

∮ EdA (top) + ∮ EdA (bottom) = Q/ε₀
Since E is constant along the surface:
E∮dA (top) + E∮dA (bottom) = Q/ε₀

2EA = Q/ε₀ (A: the area of the top/bottom)
2EA = σA/ε₀ (σ: charge density per unit area)
∴E= σ/2ε₀

Question 5

what’s Earnshaw’s theorem

Answer 5

it’s impossible to set up an electric field that will hold a charged particle in empty space

Question 6

what’s an equipotential surface

Answer 6

a surface along which has the same electric potential (ϕ) value.

the work done moving across this surface is always 0.

Electric fields are always normal to the equipotential surface

The closer together 2 surfaces are, the stronger the E-field will be

Question 7

prove Earnshaws-theorem

Answer 7

proof by falsification (assume it is possible):

take a test charge (positive). for it to be in stable equilibrium it must have an E-field pointing towards it from all directions.

take a Gaussian surface around this charge. since the E-field points inwards, ∮ E⋅dA is negative, hence there must be a negative charge contained within the surface.

This contradicts the statement ‘in empty space’