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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
C Correct answer
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, $\dfrac{F}{4}$
Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The relative permittivity of water is 81. If $\varepsilon _0$ and $\varepsilon$ are permittivities of vaccum and water respectively Then:

  1. $\varepsilon _0 = 9\varepsilon _w$
  2. $\varepsilon _0= 81\varepsilon _w$
  3. $\varepsilon _w = 9\varepsilon _0$
  4. $\varepsilon _w = 81\varepsilon _0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The relative permittivity of water id defined as:

$\varepsilon _{rel}=\dfrac{\varepsilon _w}{\varepsilon _0}$

Substituting the values:
$81=\dfrac{\varepsilon _w}{\varepsilon _0}$

So, $\varepsilon _w=81\ \varepsilon _0$

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

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of ${ 30 }^{ o }$ with each other. When suspended in a liquid of density $0.8g { cm }^{ -3 }$ the angle remains the same. If density of the material of the sphere is $1.6g { cm }^{ -3 }$ the dielectric constant of the liquid is:

  1. 2

  2. 1

  3. 4

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equilibrium in air: tan(theta) = F_e / (mg). In liquid: tan(theta) = F_e' / (mg - F_buoyant). Since theta is same, F_e/mg = F_e'/(mg - F_b). Substituting F_e' = F_e/K and F_b = V*sigma*g, we get K = rho / (rho - sigma) = 1.6 / (1.6 - 0.8) = 1.6 / 0.8 = 2.

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

The capacitance of an air filled parallel plate capacitor is $8 pF$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $80 \times 10^{-12}$ farads. 
The dielectric constant of wax is

  1. $5.0$
  2. $10.0$
  3. $8.0$
  4. $4.2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

C_air = epsilon_0 * A / d = 8 pF. C_wax = K * epsilon_0 * A / (d/2) = 2 * K * C_air = 80 pF. Thus 2 * K * 8 = 80, so K = 5.