Practice important
Dielectric Interfaces
MCQ questions and answers.
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MCQ Questions
1. A point charge q is placed at a distance a from a plane interface separating two infinite dielectric media of permittivities ε1 and ε2, with the charge in medium 1. What is the force on the charge due to polarization charges induced at the interface?
A.
F = [q²/(16πε1a²)] [(ε1 − ε2)/(ε1 + ε2)], directed away from the interface when ε1 > ε2
B.
F = [q²/(16πε1a²)] [(ε2 − ε1)/(ε1 + ε2)], directed away from the interface when ε1 > ε2
C.
F = [q²/(8πε1a²)] [(ε1 − ε2)/(ε1 + ε2)], directed away from the interface when ε1 > ε2
D.
F = [q²/(16πε2a²)] [(ε1 + ε2)/(ε1 − ε2)], directed away from the interface when ε1 > ε2
Correct Answer:
F = [q²/(16πε1a²)] [(ε1 − ε2)/(ε1 + ε2)], directed away from the interface when ε1 > ε2
Explanation:
The interface can be replaced, for the field in medium 1, by an image charge q' = q(ε1 − ε2)/(ε1 + ε2) located symmetrically across the interface. Its separation from q is 2a, so Coulomb's law gives F = qq'/(4πε1(2a)²), yielding the stated result.
2. At a boundary between two linear, isotropic dielectrics with permittivities ε1 and ε2, there is no free surface charge. If the normal electric field in medium 1 is E1n, what is the normal electric field in medium 2?
A.
E2n = (ε1/ε2)E1n
B.
E2n = (ε2/ε1)E1n
C.
E2n = E1n
D.
E2n = [(ε1 + ε2)/(ε1 − ε2)]E1n
Correct Answer:
E2n = (ε1/ε2)E1n
Explanation:
With no free surface charge, the normal component of electric displacement is continuous: D2n = D1n. Since D = εE, ε2E2n = ε1E1n, giving E2n = (ε1/ε2)E1n.
3. A uniform electric field in dielectric 1 makes an angle θ1 with the normal to a planar interface with dielectric 2. If there is no free surface charge, which relation connects the refraction angles θ1 and θ2 of the electric field?
A.
tan θ2 = (ε1/ε2) tan θ1
B.
tan θ2 = (ε2/ε1) tan θ1
C.
sin θ2 = (ε2/ε1) sin θ1
D.
tan θ2 = tan θ1
Correct Answer:
tan θ2 = (ε2/ε1) tan θ1
Explanation:
The tangential electric field is continuous, so E2t = E1t. The normal displacement is continuous, so E2n = (ε1/ε2)E1n. Therefore tan θ2 = E2t/E2n = (ε2/ε1)(E1t/E1n) = (ε2/ε1)tan θ1.
4. A charge q is embedded in dielectric 1 of relative permittivity 4 and lies near a planar interface with dielectric 2 of relative permittivity 9. The charge is at distance a from the interface. What is the magnitude and direction of the force caused by the induced polarization charge?
A.
q²/(208πε0a²), toward the interface
B.
q²/(208πε0a²), away from the interface
C.
5q²/(208πε0a²), toward the interface
D.
q²/(104πε0a²), toward the interface
Correct Answer:
q²/(208πε0a²), toward the interface
Explanation:
Here ε1 = 4ε0 and ε2 = 9ε0. The force is F = q²(ε1 − ε2)/[16πε1a²(ε1 + ε2)] = −q²/(208πε0a²). The negative sign indicates attraction toward the interface.
5. A point charge q is located in dielectric 1 near a planar interface with dielectric 2. What image charge must be used to calculate the potential in dielectric 1?
A.
q' = q(ε1 − ε2)/(ε1 + ε2), located at the mirror point in dielectric 2
B.
q' = q(ε2 − ε1)/(ε1 + ε2), located at the mirror point in dielectric 2
C.
q' = 2qε1/(ε1 + ε2), located at the mirror point in dielectric 2
D.
q' = q(ε1 + ε2)/(ε1 − ε2), located at the original position of q
Correct Answer:
q' = q(ε1 − ε2)/(ε1 + ε2), located at the mirror point in dielectric 2
Explanation:
The boundary conditions at a planar dielectric interface are satisfied in medium 1 by the real charge q plus an image charge q' = q(ε1 − ε2)/(ε1 + ε2) placed symmetrically across the interface.
6. For a point charge q in dielectric 1 near a planar interface with dielectric 2, what is the effective source strength appearing in the potential throughout dielectric 2?
A.
qeff = q(ε1 − ε2)/(ε1 + ε2)
B.
qeff = 2qε1/(ε1 + ε2)
C.
qeff = 2qε2/(ε1 + ε2)
D.
qeff = q(ε1 + ε2)/(2ε1ε2)
Correct Answer:
qeff = 2qε2/(ε1 + ε2)
Explanation:
In medium 2, the transmitted potential has the form φ2 = [1/(4πε2)] [2qε2/(ε1 + ε2)](1/R). Thus the effective source strength in the Coulomb form using ε2 is qeff = 2qε2/(ε1 + ε2).
7. At an interface between two dielectric media, the tangential component of the electric field is E1t in medium 1. In electrostatics, what is the corresponding tangential component in medium 2?
A.
E2t = (ε1/ε2)E1t
B.
E2t = (ε2/ε1)E1t
C.
E2t = E1t
D.
E2t = [(ε1 − ε2)/(ε1 + ε2)]E1t
Correct Answer:
E2t = E1t
Explanation:
Because the electrostatic field is conservative, its line integral around an infinitesimal loop crossing the interface must vanish. This requires continuity of the tangential electric field: E1t = E2t.
8. The electrostatic energy associated with a point charge q at distance a from a planar interface between dielectrics ε1 and ε2 is changed by the presence of the interface. What is the interface-dependent energy contribution?
A.
U = q²(ε1 − ε2)/[16πε1a(ε1 + ε2)]
B.
U = q²(ε1 − ε2)/[8πε1a(ε1 + ε2)]
C.
U = q²(ε1 + ε2)/[16πε1a(ε1 − ε2)]
D.
U = q²(ε2 − ε1)/[16πε2a(ε1 + ε2)]
Correct Answer:
U = q²(ε1 − ε2)/[16πε1a(ε1 + ε2)]
Explanation:
The image potential at the real charge is φimage = q'/(4πε1(2a)), where q' = q(ε1 − ε2)/(ε1 + ε2). The electrostatic energy is one-half qφimage, giving the stated expression.
9. A dielectric interface has a free surface charge density σf. If the unit normal is directed from medium 1 to medium 2, which boundary condition is correct for the normal component of electric displacement?
A.
D2n − D1n = 0
B.
D2n − D1n = σf
C.
D2n + D1n = σf
D.
ε1D2n − ε2D1n = σf
Correct Answer:
D2n − D1n = σf
Explanation:
Applying Gauss's law to a thin pillbox crossing the interface gives the discontinuity of the normal displacement as the free surface charge density: (D2 − D1)·n = σf.
10. A dielectric 1 with relative permittivity 3 touches dielectric 2 with relative permittivity 6. At the interface there is no free surface charge, and the electric field in dielectric 1 has components E1n = 12 V/m normal to the interface and E1t = 5 V/m tangential to it. What is the magnitude of the electric field in dielectric 2?