Electromagnetism - Capacitor

A parallel-plate capacitor is connected to a battery. V₀ is the potential difference between the plates, Q₀ is the charge on the positive plate, E₀ the magnitude of the electric field, and D₀ the magnitude of the displacement vector. The original vacuum between the plates is filled with a dielectric and then the battery is disconnected. If the corresponding electrical parameters for the final state of the capacitor are denoted by a subscript f, which of the following is true?

A. V_f  > V₀ 
B. V_f  < V₀
C. Q_f  = Q₀
D. E_f  > E₀
E. D_f  > D₀

(GR9277 #88)

Solution:

(A), (B) FALSE
The battery is disconnected after the plates is filled with a dielectric
V_f  = V₀ = constant

(C) FALSE
Charge: Q = CV

V_f  = V₀ and when dielectric is inserted: C_f  = κC₀
Q_f  = C_f V_f  = κC₀V₀ = κQ₀

(D) FALSE
Potential: V = Ed
V_f  = V₀ and d is not changed
E_f  = E₀

(E) TRUE
Displacement vector in empty space: D₀ = ε₀E
with dielectric constant: D_f  = κε₀E
D_f  > D₀

Answer: E

Special Relativity - Velocity of Light

For blue light, a transparent material has a relative permittivity (dielectric constant) of 2.1 and a relative permeability of 1.0. If the speed of light in vacuum is c, the phase velocity of blue light in an unbounded medium of this material is

A. √3.1 c
B. √2.1 c
C. c/√1.1
D. c/√2.1
E. c/√3.1

(GR8677 #03)

Solution:

Permittivity, ϵ = 2.1 ϵ₀
Permeability, μ = μ₀
Speed of light in vacuum:
The velocity of light in medium:  

Answer: D

Electromagnetism - Dielectric

A dielectric of dielectric constant K is placed in contact with a conductor having surface charge density σ, as shown above. What is the polarization (bound) charge density σ_p on the surface of the dielectric at the interface between the two materials? 

A. σK / (1−K)
B. K / (1 + K)
C. σK
D. σ(1 + K) / K
E. σ(1 − K) / K

(GR8677 #54)

Solution:

Polarization, P = ɛ₀χ_eE = ɛ₀(K − 1)E
Thus, K  1.
If K = 1 → P = 0 and σ_p = 0 → no polarized dielectric

Check answers for K = 1→ σ_p = 0

(A) σ_p = σK/(1 − K) = ∞ → FALSE
(B) σ_p = K/(1 + K) = 1/2 → FALSE
(C) σ_p = σK = σ → FALSE
(D) σ_p = σ(1 + K)/K = 2σ → FALSE
(E) σ_p = σ(1 − K)/K = 0 → TRUE

Answer: E 


Alternative Solution:

In dielectric, E = σ/ɛ = σ/ɛ₀K
Polarization, P = ɛ₀(K − 1)E
σ_p = ɛ₀(K − 1)(σ/ɛ₀K)
= (K − 1)(σ/K)
= σ(1 − K)/K

Electromagnetism – Dielectric

An infinite slab of insulating material with dielectric constant K and permittivity ɛ = Kɛ₀ is placed in a uniform electric field of magnitude E₀. The field is perpendicular to the surface of the material. The magnitude of the electric field inside the material is 

A. E₀/K
B. E₀/Kɛ₀ 
C. E₀
D. Kɛ₀E₀
E. KE₀

(GR0177 #95)

Solution:

In vacuum, E₀ = σ/ɛ₀
In dielectric, E = σ/ɛ
where ɛ = Kɛ₀
Thus, E = σ/Kɛ₀ = E₀/K

Answer: A