Match the following lists in terms of relative permittivity :List – I List – II a. Air i. 25 b. Alcohol (Ethyl) ii. 4 to 7 c. Glass iii. 80 d. Fresh water iv. 1
a-iv, b-i, c-ii, d-iii
Anchor the two extremes first.
a → iv. Air has εr = 1. Strictly it is 1.0006, so for every practical purpose air behaves like free space. This is why antenna and transmission-line calculations in air use \(c=1/\sqrt{\mu_0\epsilon_0}\) unchanged. Every option except 3 agrees on this, so it merely confirms the direction of the table.
d → iii. Fresh water has εr ≈ 80. This famously large value comes from the water molecule's permanent dipole moment: the molecules rotate to align with an applied field, and orientational polarisation is far stronger than the electronic polarisation available to non-polar materials. It is also why water heats in a microwave oven — and why the value falls sharply above a few GHz, as the molecules can no longer follow the field.
b → i. Ethyl alcohol has εr ≈ 25. Alcohol is also a polar molecule, so its permittivity is high, but its hydroxyl group is attached to a bulky ethyl chain that hinders rotation, so it reaches only about a quarter of water's value. Placing it between glass and water is the physically sensible ordering.
c → ii. Glass has εr = 4 to 7. A range rather than a single number, because "glass" covers many compositions — borosilicate near 4, lead glass nearer 7. Solid non-polar and weakly polar dielectrics typically sit in this 2–10 band, alongside mica (6), rubber (3) and paper (2–4).
Assemble. a-iv, b-i, c-ii, d-iii — option 4.
The physical ordering to remember. Gases ≈ 1, non-polar solids 2–10, polar liquids tens, water about 80, ferroelectric ceramics such as barium titanate in the thousands. Ranking by polarity rather than trying to memorise numbers gets this question right every time: air (none) < glass (weak) < alcohol (polar, hindered) < water (strongly polar).
Hence, the correct matching is a-iv, b-i, c-ii, d-iii.
Similar Questions
For a uniform line charge of 8 nc/m lying along the z axis, the electric field at a radius of 3 m from the uniform line is given by :
Given below are two statements :
Statement I : When an electron is brought near the surface of a metal in any dielectric medium, an image charge of opposite polarity is included inside the metal at the same distance from the surface as that of the electron in the dielectric medium.
Statement II : In the case of metal-semiconductor junction, an electron in the semiconductor also creates an image charge in the metal which decreases the energy barrier height.
In the light of the above statements, choose the correct answer from the options given below :
Important Questions from Electrostatics
Three point charges q are placed at the corners of an equilateral triangle. Another point charge −Q is placed at the centroid of the triangle. If the force on each of the charges q vanishes, then the ratio Q/q is
The components of the electric field, in a region of space devoid of any charge or current sources, are given to be E i= a i+ Σ j=1,2,3 bij xj , where a iand b ij are constants independent of the coordinates. The number of independent components of the matrix b ij , is
Whenever a conductor cuts magnetic flux, an e.m.f. is induced in that conductor. This phenomenon is according to
The value of electric field E at a point in Electric field of a point charge can be calculated using:
An inductor of 3.3mH with a series resistance of 12.5 ohms is connected to a 5V dc supply. When the supply is switched off, the circuit current decay to zero in 60 microseconds. What is the value of back e.m.f. generated?
