Polarisation of a dielectric is defined as
Answer: A
It is a vector, measured in coulombs per square metre, the same units as flux density. Its divergence gives the bound volume charge density with a change of sign, and its normal component at a surface gives the bound surface charge there.
Please log in or register to report this question.
A dielectric placed in an external field acquires
Answer: B
Correct answer: bound charge on its surfaces but no net charge overall
The field aligns or induces molecular dipoles, and within the material the head of one dipole cancels the tail of its neighbour, so charge appears only where that cancellation fails - at the two surfaces, positive on one and negative on the other. The magnitudes are equal, so the material remains neutral overall, and the charge is bound because it cannot leave the molecules that carry it.
Please log in or register to report this question.
The field inside a dielectric placed in an external field is
Answer: C
The bound surface charge produces a field opposing the applied one, so the resultant within the material is reduced by the factor of the relative permittivity. This reduction is exactly what makes a dielectric increase the capacitance of an arrangement.
Please log in or register to report this question.
Electric susceptibility relates the polarisation to
Answer: D
Correct answer: the field intensity, through the permittivity of free space
\[ \mathbf{P} = \chi_e\,\epsilon_0\mathbf{E}, \] which is the definition of the electric susceptibility. It connects to the permittivity through \( \epsilon_r = 1 + \chi_e \), so a vacuum has zero susceptibility and unit relative permittivity - and the polarisation is driven by \( \mathbf{E} \), the field actually present in the material.
Please log in or register to report this question.
The relative permittivity of a material exceeds the susceptibility by
Answer: A
From \( \mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} \) and the definition of susceptibility, \( \epsilon_r = 1 + \chi_e \). A vacuum has zero susceptibility and unit relative permittivity, which is the consistency check on the relation.
Please log in or register to report this question.
The relative permittivity of a dielectric is always
Answer: B
Polarisation always acts to oppose the applied field within the medium, which corresponds to a positive susceptibility and therefore a relative permittivity above one. Values range from about 1.0006 for air to several thousand for some ceramics.
Please log in or register to report this question.
The constitutive relation for a linear isotropic dielectric is
Answer: C
Correct answer: \( \mathbf{D} = \epsilon_0\epsilon_r\mathbf{E} \)
Both factors are needed: \( \epsilon_0 \) carries the units and \( \epsilon_r \) describes the material, so their product is the absolute permittivity. Equivalently \( \mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} \), which shows what the relative permittivity is doing - adding the polarisation to the vacuum term. The first option omits \( \epsilon_0 \) and is dimensionally wrong.
Please log in or register to report this question.
At the interface between two dielectrics with no free surface charge, the quantity that is continuous is
Answer: D
Gauss's law applied to a thin box straddling the surface gives equal normal flux on the two sides when no free charge is present. The normal field intensity is therefore discontinuous, in inverse proportion to the permittivities.
Please log in or register to report this question.
The component of the electric field that is always continuous across a dielectric boundary is
Answer: A
Taking the line integral of E round a thin rectangle straddling the boundary and shrinking it gives equal tangential components, since the static field is conservative. The tangential flux density is correspondingly discontinuous.
Please log in or register to report this question.
A field crossing from a dielectric of low permittivity into one of high permittivity is bent
Answer: B
The tangential field is preserved while the normal field falls in the denser medium, so the resultant tilts further from the normal. The relation \( \tan\theta_1/\tan\theta_2 = \epsilon_1/\epsilon_2 \) is the electrostatic analogue of Snell's law.
Please log in or register to report this question.
