Tag: electrostatics

Quizzes (15) Questions (156)

Questions Related to electrostatics

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The value of relative permittivity of air is

  1. $8.854\times { 10 }^{ -12 }{ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }$
  2. $9\times { 10 }^{ 9 }{ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }$
  3. $1$
  4. $8.854\times { 10 }^{ 12 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The relative permittivity of a material is ratio of its (absolute) permittivity to the permittivity of vacuum. For air it is almost 1. 

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

A plane electromagnetic wave in a non magnetic dielectric medium is given by $\bar{E} = \bar{E _0} ( 4 \times 10^{-7} \times - 50 t)$ with distance being in meter and time in seconds. The dielectric constant of the medium is:

  1. 2.4

  2. 5.8

  3. 8.2

  4. 4.8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation of wave 

$\vec{E} = \vec{E} _0 (4 \times \vec{10} x - 50 t)$
Comparing with general equation of wave,
$\vec{E} = \vec{E} _0 (kx - wt),$
we get,
$w = 50\ rad/s.$   $k = 4 \times 10^{-7} m^{-1}$
Thus velocity of wave, 
$v = \dfrac{w}{k} = \dfrac{50}{4 \times 10^{-7}} = 1.25 \times 10^8 m/s$
so, refractive index of medium,
$\mu = \dfrac{e}{v} = \dfrac{3 \times 10^8}{1.25 \times 10^8} = 2.4$
using, 
$u^2 = km.ke$     [km and ke are magnetic and dielectric di - constants]
as medium is non diamagnetic, $km = 1$
$\Rightarrow ke = \mu^2 = (2.4)^2 \Rightarrow ke = 5.76$

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

Two charges placed in air repel each other by a force of $10^{-4}N$. When  oil is introduced between the charges, the force becomes $2.5 \times 10^{-5} N$. The dielectric constant of oil is: 

  1. $2.5$
  2. $0.25$
  3. $2.0$
  4. $4.0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The dielectric constant K is defined as the ratio of the force in vacuum (or air) to the force in the medium. K = F_air / F_medium = 10^-4 / (2.5 * 10^-5) = 10 / 2.5 = 4.

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The dielectric constant $k$ of a medium can be defined as $k=M/N$, where $N$ is the capacity of a parallel plate capacitor. When the space between plates is filled with air, $M$ is ?

  1. the charge of the capacitor

  2. the capacity of a parallel plate capacitor with a dielectric between the plates

  3. the dielectric intensity in the space between the plates

  4. the P.D. across the plates of the condenser, with a dielectric between them
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The dielectric constant k of a medium is defined as the ratio of the capacitance of a capacitor filled with that medium (M) to the capacitance of the same capacitor with a vacuum or air (N). Therefore, M represents the capacity with the dielectric between the plates.

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

$64$ small drops of mercury each of radius $r$ and change $q$ coalesce to from a big drop. The ratio of the surface charge density of each small drop with that of big drop is:

  1. $4:1$
  2. $1:4$
  3. $1:64$
  4. $64:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Formulae:

$\dfrac{\sigma _{small} }{\sigma _{big} }=\dfrac{q}{Q} \times \dfrac{R^2}{r^2}$

$=\dfrac{q}{nq}\times \dfrac{(n^{\frac{1}{3}}r)^2}{r^2}$

$=n^{-\frac{1}{3}}$

$=64^{-\frac{1}{3}}$

$=\dfrac{1}{4}$

hence the ratio is $1:4$
Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The force of attraction between two charges separated by certain distance in air is F1. If the space between the charges is completely filled with dielectric of constant 4 the force becomes F2. If half of the distance between the charges is filled with same dielectric the force between the charges is F3. Find F1:F2:F3 is

  1. 16 : 9 : 4

  2. 9 : 36 : 16

  3. 4 : 1 : 2

  4. 36 : 9 :16
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

Two fixed charges separated by a distance $d$ experience a force $F$. A dielectric medium of thickness $\dfrac{d}{4}$ and dielectric constant $4$ is introduced in the space between them. Find the new force between the charges.

  1. $\dfrac{F}{4}$
  2. $\dfrac{F}{3}$
  3. $F$
  4. $\dfrac{16F}{25}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The force between the 2 charged particles is inversely proportional to the permittivity.

Therefore if the permittivity increases by 4 times, then obviously the force decreases by 4 times.

Therefore the new force is given by, $\dfrac{F}{4}$