physics : class12 Gauss Law in Dielectrics : Insulators : Electrostatics MCQ Questions and Answers
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Gauss Law in Dielectrics
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MCQ Questions
1. An infinite plane sheet carries a free surface charge density σ_f and lies at the interface between two linear, homogeneous dielectrics of permittivities ε_1 and ε_2. Assuming the electric displacement is normal to the interface and the fields vanish at infinity, what is the ratio E_1/E_2 of the electric-field magnitudes on the two sides?
A.
ε_1/ε_2
B.
ε_2/ε_1
C.
1
D.
(ε_1 + ε_2)/(2ε_1ε_2)
Correct Answer:
ε_2/ε_1
Explanation:
Gauss's law for electric displacement gives D_1 = D_2 = σ_f/2 in magnitude on the two sides. Since E_i = D_i/ε_i, E_1/E_2 = ε_2/ε_1.
2. A point free charge q is placed at the center of a spherical dielectric of radius R and permittivity ε. The dielectric is surrounded by vacuum. What is the total bound charge induced on the dielectric surface r = R?
A.
q(ε - ε_0)/ε
B.
q(ε_0 - ε)/ε_0
C.
q(ε - ε_0)/ε_0
D.
0
Correct Answer:
q(ε - ε_0)/ε
Explanation:
Inside the dielectric, Gauss's law for D gives D = q/(4πr²), so E = q/(4πεr²). Thus P = (ε - ε_0)E. The total bound surface charge is 4πR²P(R) = q(ε - ε_0)/ε.
3. A closed Gaussian surface lies entirely within a homogeneous dielectric and encloses a free charge q_f as well as an arbitrary distribution of bound charge. Which relation is always valid?
A.
∮E·dA = q_f/ε
B.
∮D·dA = q_f
C.
∮D·dA = (q_f + q_b)/ε_0
D.
∮P·dA = q_f
Correct Answer:
∮D·dA = q_f
Explanation:
Gauss's law in dielectrics is ∮D·dA = Q_free,enclosed. Bound charges are incorporated into D and do not appear explicitly on the right-hand side.
4. A dielectric slab of permittivity ε occupies the region between two infinite parallel planes. It contains a uniform free volume charge density ρ_f and is symmetric about x = 0. What is the electric field at x = a inside the slab?
A.
ρ_f a/(2ε)
B.
ρ_f a/ε
C.
ρ_f/(2εa)
D.
ρ_f a/(2ε_0)
Correct Answer:
ρ_f a/ε
Explanation:
Using a pillbox symmetric about x = 0, the displacement satisfies D_x(a) = ρ_f a because the enclosed free charge per unit area is ρ_f a. Therefore E(a) = D(a)/ε = ρ_f a/ε.
5. A parallel-plate capacitor is completely filled with a linear dielectric of relative permittivity κ. The free charge magnitude on each plate is Q. What is the magnitude of the bound charge induced on either dielectric surface adjacent to a plate?
A.
Q/κ
B.
Q(κ - 1)/κ
C.
Q(κ + 1)/κ
D.
Q(1 - κ)
Correct Answer:
Q(κ - 1)/κ
Explanation:
The free surface charge gives D = Q/A, while E = D/(κε_0). Hence P = ε_0(κ - 1)E = [(κ - 1)/κ](Q/A). Multiplying by area gives the bound charge magnitude Q(κ - 1)/κ.
6. A long coaxial arrangement has a line of free charge density λ on its axis and is filled between radii a and b with a radially varying permittivity ε(r). What is the electric displacement magnitude at radius r, where a < r < b?
A.
λ/[2πrε(r)]
B.
λ/[2πrε_0]
C.
λ/(2πr)
D.
λ/[4πr²]
Correct Answer:
λ/(2πr)
Explanation:
A cylindrical Gaussian surface of radius r and length L encloses free charge λL. Therefore D(2πrL) = λL, giving D(r) = λ/(2πr), independent of the radial variation of ε.
7. A spherical dielectric of permittivity ε contains a uniform free volume charge density ρ_f throughout its volume. For a point inside the sphere, what is the bound volume charge density ρ_b?
A.
−ρ_f(ε − ε_0)/ε
B.
ρ_f(ε − ε_0)/ε
C.
−ρ_fε/ε_0
D.
0
Correct Answer:
−ρ_f(ε − ε_0)/ε
Explanation:
Gauss's law gives D = ρ_f r/3 and E = ρ_f r/(3ε). Hence P = (ε − ε_0)E. Since ∇·P = (ε − ε_0)ρ_f/ε, the bound volume charge is ρ_b = −∇·P = −ρ_f(ε − ε_0)/ε.
8. At an interface between two dielectrics, the free surface charge density is σ_f. The normal components of electric displacement immediately on the two sides are D_{1n} and D_{2n}, with the unit normal directed from medium 1 to medium 2. Which boundary relation follows from Gauss's law in dielectrics?
A.
D_{2n} + D_{1n} = σ_f
B.
D_{2n} - D_{1n} = σ_f
C.
ε_2D_{2n} - ε_1D_{1n} = σ_f
D.
D_{2n}/ε_2 - D_{1n}/ε_1 = σ_f
Correct Answer:
D_{2n} - D_{1n} = σ_f
Explanation:
Applying Gauss's law for D to a thin pillbox crossing the interface gives the discontinuity of the normal component as the free surface charge density: (D_2 − D_1)·n = σ_f.
9. A point free charge q is placed at the center of a spherical cavity inside an infinite homogeneous dielectric of permittivity ε. For a spherical Gaussian surface lying entirely in the dielectric and centered on the charge, which electric field magnitude is correct?
A.
q/(4πε_0r²)
B.
q/(4πεr²)
C.
q/(4π(ε − ε_0)r²)
D.
q/(4πε_0εr²)
Correct Answer:
q/(4πεr²)
Explanation:
The Gaussian surface encloses free charge q, so D = q/(4πr²). Since the surface lies in the dielectric, E = D/ε = q/(4πεr²).
10. A dielectric sphere of radius R and permittivity ε contains a point free charge q at its center. What is the electric displacement flux through any spherical surface of radius r > R in the surrounding vacuum?
A.
qε_0/ε
B.
qε/ε_0
C.
q
D.
0
Correct Answer:
q
Explanation:
Gauss's law for D counts only enclosed free charge. Any spherical surface outside the dielectric encloses the free charge q, regardless of the bound charges induced in the dielectric, so the D-flux is q.
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