Tag: electrostatics

Questions Related to electrostatics

 The capacitance of a capacitor

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

  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 

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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 

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

  1. expanding the electrode areas 

  2. reducing the gap between the electrodes

  3. expanding the gap between the electrodes

  4. Both A and B

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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.

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

  1. $2$

  2. $4$

  3. $6$

  4. $3$

Show answer
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?

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

  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

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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 ________ $m/{s}^{-1}$

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

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

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

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

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

Show answer
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

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

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

Show answer
Correct Option: A

Two charges $+Q$ and $-2Q$ are located at points $A$ and $B$ on a horizontal line as shown in the diagram.
The electrical field is zero at a point which is located at finite distance :

physics electrostatics field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. On the perpendicular bisector of $AB$

  2. Left of $A$ on the line

  3. Between $A$ and $B$ on the line

  4. Right of $B$ on the line

Show answer
Correct Option: B
Explanation:

$ E _1 $ = Electrical field due to $+Q$
$E _2$ = Electrical feild due to $-2Q$
There resultant is $0$ at this point 

Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding . The charges on the two rings are + q and -q . The potential difference between the centres of the two rings is

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $\frac {{ Q.R }\quad} {\quad { 4\pi }{ \varepsilon } _{ 0 }{ d }^{ 2 }}$

  2. $\frac { Q }{ 2\pi { \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { R }^{ 2 }+{ d }^{ 2 } } } ]$

  3. $\frac { Q }{ { 4\pi \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { { R }^{ 2 } }+{ { d }^{ 2 } } } } ]$

  4. 0

Show answer
Correct Option: A
Explanation:

Two thin wire ring each having a radius $=R$

distance $=d$
two rings are $+q$ and $-q$.
$PD=\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \dfrac { QR }{ { d }^{ 2 } } $
Now in this question, we get the solution.

what is the potential difference between two points, if 2J of work must be done to move a 4 mC charge from one point to another is:

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. 50 V

  2. 500 V

  3. 5 V

  4. 5000 V

Show answer
Correct Option: B
Explanation:

Answer is B.

The total work done = energy transferred.
so, we might see the equation energy = voltage x charge, E = V * Q, written as, 
work = voltage x charge, W = V * Q.
In this case, the charge is 4 mC, that is, 0.004 C and work done is 2 J.
Therefore, V=W/Q = 2/0.004 = 500 V.
Hence, the potential difference between two points if 2 J of work must be done to move a 4 mC charge from one point to another is 500 V.

A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is_______

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. zero

  2. positive

  3. negative

  4. zero if the charge Q is at the center and nonzero otherwise.

Show answer
Correct Option: A
Explanation:

The net displacement round one complete circle is 0.

So, the work done is 0.