Tag: electrostatic potential and capacitance

Questions Related to electrostatic potential and capacitance

Which of the following statement is correct regarding dielectric?

  1. Dielectric constant is the numerical value which signifies the extent of electric field decreased passing through it.

  2. Conductor can be considered as dielectric having dielectric constant as infinite

  3. The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space.

  4. None of the above


Correct Option: A,B,C

Which of the material can be used as a dielectric?

  1. Ceramics

  2. Wood

  3. Copper

  4. Aluminium


Correct Option: A,B
Explanation:

we know that only non conductor can be used as a dilectric.So only A and B are correct option.

Dielectric constant, property of an electrical insulating material (a dielectric) equal to

  1. the ratio of the capacitance of a capacitor filled with the given material to the capacitance of an identical capacitor in a vacuum without the dielectric material.

  2. the ratio of the capacitance of a capacitor filled with the given material to the capacitance of an identical capacitor in a vacuum with the dielectric material.

  3. the capacitance of a capacitor filled with the given material 

  4. none  of the above 


Correct Option: A
Explanation:

The capacitance of a parallel plate capacitor filled with the given material of dielectric constant $K$ is given by:

              $C=\dfrac{K\varepsilon _{0}A}{d}$ .....................eq1
And the capacitance of the same capacitor without the given material is given by:
              $C'=\dfrac{\varepsilon _{0}A}{d}$  .....................eq2
Dividing eq1 by eq2,
              $C/C'=K$

Which of the following can be used as dielectric?

  1. Mineral oil

  2. Electrets

  3. Sulfur hexafluoride

  4. None of the above


Correct Option: A,B,C
Explanation:

We know that dielectric is a medium or substance with a dielectric property; an insulator.So, from the option Mineral oil,Electrets and also Sulfur hexafluoride can be used as dilectric.

Therefore, A,B and C are correct option.

 The capacitance of a capacitor

  1. filled with a dielectric is lesser than it would be in a vacuum.

  2. filled with a dielectric is greater than it would be in a vacuum.

  3. filled with a dielectric is same as it would be in a vacuum.

  4. none of the above 


Correct Option: B
Explanation:

The capacitance of a parallel plate capacitor filled with dielectric of dielectric constant $K$ is,

           $C=\dfrac{K\varepsilon _{o}A}{d}$  ..................eq1
and the capacitance of a parallel plate capacitor without dielectric in a vacuum is,
           $C'=\dfrac{\varepsilon _{o}A}{d}$  ...................eq2
Dividing eq1 by eq2 , we get
            $C=KC'$
as $K>1$ , therefore $C>C'$ therefore the capacitance of a capacitor filled with a dielectric is greater than it would be in a vacuum.

The voltage can be increased, but electric breakdown will occur if the electric field inside the capacitor becomes too large. The capacity can be increased by 

  1. expanding the electrode areas 

  2. reducing the gap between the electrodes

  3. expanding the gap between the electrodes

  4. Both A and B


Correct Option: D
Explanation:

The capacitance of an air filled parallel plate capacitor is given by:

            $C=\dfrac{\varepsilon _{0}A}{d}$  ..........................eq1
where $A=$ area of each plate (electrode),
            $d=$ distance between plates (electrodes)
from eq1 ,
           $C\propto A$
It is clear that capacity C can be increased by expanding the electrodes area,
and     $C\propto 1/d$ ,
hence, capacity C can be increased by reducing the gap between the electrodes.

A dielectric slab of thickness $6 cm$ is placed between the plates of a parallel plate capacitor. If the distance between plates is reduced by $4 cm$, the capacity of the capacitor remains the same. Find the dielectric constant of the medium.

  1. $2$

  2. $4$

  3. $6$

  4. $3$


Correct Option: D
Explanation:
 for a parallel plate capacitor 
             C = ε K   A / d
It is possible that we have air/vacuum and then a medium in between the parallel plates.  Initially there is a medium of dielectric constant K and of thickness d.

    Let  K1 be the dielectric constant of the slab.  
     let d1 be the thickness of slab = 6cm.

Formula for capacity with multiple media in between the plates is 
             C =  ε A / [ d1/K1 + d2 / K ]
          total gap =  d1 + d2 = d + 4 cm          =>  d2 = d + 4 cm - 6 cm = d - 2
 
         d / K =  d1/ K1 + d2 / K 
         d/ K = 6 / K1  + (d-2)/K

          d - (d-2)  = 6 K / K1 
                K1 = 3 K
  If there was air originally in between the plates,  K = 1,  then answer is 3.

A parallel plate condenser is charged to $100$ volt. In this context which one of the following statements is true?

  1. The two plates attract each other

  2. There is to force between the plates

  3. The two plates of the condenser repel each other

  4. The force between the plates can be attractive or repulsive depending upon the nature of the dielectric material between the plates


Correct Option: D

At room temperature, if the relative permittivity of water be $80$ and the relative permeability be $0.0222$, then the velocity of light in water is ________ $m/{s}^{-1}$

  1. $3\times { 10 }^{ 8 }\quad $

  2. $2.25\times { 10 }^{ 8 }$

  3. $2.5\times { 10 }^{ 8 }$

  4. $3.5\times { 10 }^{ 8 }$


Correct Option: B
Explanation:

Formula 

$V=\cfrac{1}{\sqrt{\mu _0\epsilon _0}}\V^1=\cfrac{1}{\mu _r\epsilon _r}\ \Rightarrow \cfrac{V}{V^1}=\sqrt{\cfrac{\mu _r\epsilon _r}{\mu _0\epsilon _0}}$
$\Rightarrow\cfrac{V^1}{V}=\sqrt{\cfrac{\mu _0}{}{\mu _r}\times\cfrac{\epsilon _0}{\epsilon _r}}=\sqrt{\cfrac{1}{80}\times\cfrac{1}{0.0222}}\ \Rightarrow V^1=\sqrt{\cfrac{1}{80}\times\cfrac{1}{0.0222}}\times V\ \quad=\sqrt{\cfrac{1}{80}\times\cfrac{1}{0.022}}\times3\times10^8\ \quad=2.249\times 10^8\ \quad\approx2.22\times10^8m/s$

Surface charge density of a thin disc having radius $R$ varies with distance from centre as $\sigma =\sigma _{0}\dfrac {R}{r}(r\neq 0) $ then total charge of disc is

  1. $2\pi \sigma _{0}R^{3}$

  2. $\sqrt {2}\pi \sigma _{0}R^{2}$

  3. $2\sqrt {2}\pi \sigma _{0}R^{2}$

  4. $2\pi \sigma _{0}R^{2}$


Correct Option: A