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What is relative permittivity?

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प्रश्न

What is relative permittivity?

दीर्घउत्तर
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उत्तर

  1. Relative permittivity or dielectric constant is the ratio of the absolute permittivity of a medium to the permittivity of free space. It is denoted as K or εr.
    i.e., K or ε= `ε/(ε_0)`
  2. It is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two-point charges when placed at the same distance in the given medium.
    i.e., K or ε = `"F"_"vacuum"/"F"_"medium"`
  3. It is also called a specific inductive capacity or dielectric constant.
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पाठ 10: Electrostatics - Exercises [पृष्ठ २०६]

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बालभारती Physics [English] Standard 11 Maharashtra State Board
पाठ 10 Electrostatics
Exercises | Q 2. (ix) | पृष्ठ २०६

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Two equal balls with equal positive charge 'q' coulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two?


Four charges +q, −q, +q and −q are to be arranged respectively at the four corners of a square ABCD of side 'a'.
(a) Find the work required to put together this arrangement.
(b) A charge q0 is brought to the centre of the square, the four charges being held fixed. How much extra work is needed to do this ?


Write any two important points of similarities and differences each between Coulomb's law for the electrostatic field and Biot-Savart's law of the magnetic field ?


Plot a graph showing the variation of coulomb force (F) versus ,`(1/r^2)` where is the distance between the two charges of each pair of charges: (1 μC, 2 μC) and (2 μC, − 3 μC). Interpret the graphs obtained.


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Four equal charges of 2.0 × 10−6 C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb's force experienced by one of the charges due to the other three. 


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A particle A with a charge of 2.0 × 10−6 C and a mass of 100 g is placed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, with the same charge and mass, be placed on the incline so that it may remain in equilibrium? 


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Two equal charges, 2.0 × 10−7 C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process?


A force F acts between sodium and chlorine ions of salt (sodium chloride) when put 1 cm apart in air. The permittivity of air and dielectric constant of water are `epsilon_0` and K respectively. When a piece of salt is put in water, electrical force acting between sodium and chlorine ions 1 cm apart is ____________.


The electric force acting between two point charges kept at a certain distance in vacuum is 16 N. If the same two charges are kept at the same distance in a medium of dielectric constant 8, the electric force acting between them is ____________ N.


Polarised dielectric is equivalent to ______.


Two positive charges ______.


The unit of charge is ______.


The law, governing the force between electric charges is known as ______.

SI unit of permittivity of free space is ______.


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The S.I unit of electric permittivity is


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There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by

F = `(Qq)/r^2 hatr`

where the distance r is measured in cm (= 10–2 m), F in dynes (= 10–5 N) and the charges in electrostatic units (es units), where 1 es unit of charge = `1/([3]) xx 10^-9 C`

The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 × 108 m/s. An approximate value of c then is c = [3] × 108 m/s.

(i) Show that the coloumb law in cgs units yields

1 esu of charge = 1 (dyne)1/2 cm.

Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L.

(ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives

`1/(4pi ∈_0) = 10^-9/x^2 (N*m^2)/C^2`

With `x = 1/([3]) xx 10^-9`, we have `1/(4pi ∈_0) = [3]^2 xx 10^9 (Nm^2)/C^2`

or, `1/(4pi ∈_0) = (2.99792458)^2 xx 10^9 (Nm^2)/C^2` (exactly).


Two point charges +2 C and +6 C repel each other with a force of 12 N. If a charge of -4 C is given to each of these charges, then the force now is ______.


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