derivations

Question 1

derive the E-field from a ring of charge, q, a distance ,d, away from its central axis

Answer 1

a ring of charge, of radius a. the point is a distance d away from the central axis. dl is the distance around the ring

E = ∫ dE
dE = dQ/4πε₀r²
dQ = Q/2πa * dl = λ dl

the up and down components cancel, so only the components horizontally (chosen as the î direction) exist.

dEₓ = |dE|*cos(θ)
cos(θ) = d/r

E = ∫dEₓ = ∫ [λ/4πε₀r² * d/r] dl
= ∫ λd/4πε₀r³ dl (limits are 0 and 2πa)
= d/4πε₀r³ * ∫ λdl (∫ λdl = q)
= qd/4πε₀r³

r² = a² + d²

hence: E = qd/4πε₀ * 1/(a²+d²)^3/2

Question 2

derive the capacitance per unit length for an infinite cylindrical capacitor

Answer 2

the capacitor is a hollow cylinder with thick walls (doughnut-shaped prism)
C/l = Q/(l*Δϕ) (Q = λl)
C/l = λ/Δϕ

to find Δϕ you need to find E:
Δϕ = ∫ E⋅dl (between the limits of the inner radius and out radius of the capacitor)

E is found with Gauss’ law, treating the capacitor as an infinite wire
so E = λ/2πrε₀

hence Δϕ = λ/2πε₀ * ln(rₗ/rₛ)

sub into C/l equation:
C/l = 2πε₀ / ln(rₗ/rₛ)

Question 3

derive the electric field at a point inline with a dipole

Answer 3

‘d’ is the separation of the dipoles and ‘r’ is the distance from d/2

E = E₊ + E₋ = Q/(4πε₀* r₁²) + Q/(4πε₀* r₂²)

r₁ = r -d/2
r₂ = r + d/2

1/r₁² = 1/(r -d/2)² = 1/[r² *(1-d/2r)² ]
expand (1-d/2r)⁻² to get ≈ 1 + d/r
so 1/r₁² = 1/r² *(1+dr)
hence 1/r₂² = 1/r² *(1-dr)

E = Q/4πr²ε₀ *[ (1+d/r) - (1-d/r)] = 2Qd/4πr³ε₀

Question 4

derive the electric field perpendicular to a dipole

Answer 4

‘d’ is the separation of the dipoles and ‘r’ is the distance from d/2. ‘α’ is the angle between r and r₁ and/or r and r₂

the horizontal components will cancel
the vertical components will add

E= [2Q*sin(α) ] / 4πε₀r²

if r&raquo_space; d then α is small so α ≈ sin(α) = d/2r

∴ E = Qd/4πε₀r³ (denoted in the theta hat direction)

Question 5

derive the energy stored in a parallel plate capacitor and hence the energy density

Answer 5

start with 2 uncharged parallel plates.

charge dq is added.
this requires work dw= Φdq (this work is stored in the E-field)

Φ= q/C

W= ∫q/C dq = Q²/2C = ½ C Φ²

for a parallel plate capacitor:
Φ = E*d
C= ε₀A/d

∴ W = ½ * ε₀* E² * A * d
= ½ * ε₀* E² * V (V is the volume between the plates, V= Ad)

The energy density is therefore:
u= ½ * ε₀* E²

Question 6

derive the equation for the energy stored in a spherical capacitor

Answer 6

use the energy density equation u= ε₀E²/2

sub E = Q/4πε₀r²

then do a volume integral
dV = 4πr² dr for a sphere

this will result in:
Q²/8πε₀ [-r⁻¹] with the limits as r₁ and r₂
r₁ is the radius of the inner cavity and r₂ is the radius of the whole sphere

Question 7

use the Biot-Savart law to get an equation for the B-field from a loop of current

Answer 7

draw a loop (upright) and its central axis. the radius of the loop is a. z is the distance from the loop’s centre to a point along the central axis. r is the distance from the loop to that point θ is the angle between r and the loops surface.

hence: cos(θ) = a/r and r² = a² + z²

only the ž direction of the magnetic fields doesn’t cancel.
dB (in ž direction) = |dB|cos(θ) = µ₀I/4πr² dl cos(θ)

sub in and integrate:
B = ∮ µ₀I/4πr² cos(θ) dl
= ∮µ₀I/4πr² (a/r) dl
= [ (µ₀Ia) / 4π(a² + z²)^3/2 ]* ∮dl
= (µ₀Ia²) / 2(a² + z²)^3/2 (in the ž direction)

Question 8

use the Biot-Savart law to get an equation for the B-field from an infinite wire

Answer 8

draw a line. s is the perpendicular distance to a point P. r is the distance from P to any point on the wire. α is the angle between s and r. z is the distance between the points of interception or s and r on the wire

hence: r² = s²/cos²(α) and dl = dz ž

B = µ₀I/4π * ∫ 1/r² ž x r̂ dz

ž x r̂ = cos(α) (*theta hat)
since z = s tan(α), dz = s/cos²(α) dα

B = µ₀I/4π * ∫ cos(α)/[ s²/cos²(α) ]² * s/cos²(α) dα
= µ₀I/4π * ∫ cos(α)/s dα (limits are -π/2 to π/2 for an infinite wire)
= µ₀I/2πs (*theta hat)