Physics

Electrostatics

303 Questions

Electrostatics deals with electric charges, fields, and potentials at rest. It is a crucial topic for physics sections in engineering and civil services competitive examinations. Review these questions to build a strong understanding of Coulomb law, electric dipoles, Gauss law, and electric flux.

Electric field and potentialElectric dipole momentGauss Law applicationsCoulomb force calculationsCharge distribution on spheresEquipotential surfaces

Electrostatics Questions

Multiple choice torque on a dipole in a uniform electric field electric dipole electric charges and fields electrostatics physics

Four equal positive charges each of magnitude $q$ are placed at the respective vertices of a square of side length $l$. A point charge $Q$ is placed at the centre of the square. Then

  1. $Q$ must not be in equilibrium
  2. $Q$ must be in stable equilibrium
  3. $Q$ must be in neutral equilibrium
  4. $Q$ must be in unstable equilibrium
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The four charges create a symmetric field. A charge Q at the center is in equilibrium. If displaced, the restoring force pushes it back toward the center, indicating stable equilibrium.

Multiple choice torque on a dipole in a uniform electric field electric dipole electric charges and fields electrostatics physics

If we rotate the dipole of moment $p$ placed in an electric field $E$ from an $\theta _1$ to $\theta _2$, the work done by the external force is

  1. $pE(\cos \theta _2 - \cos \theta _1)$
  2. $pE(\cos \theta _1 - \cos \theta _2)$
  3. $pE(\sin \theta _2 - \sin \theta _1)$
  4. $pE(\sin \theta _1 - \sin \theta _2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given dipole of dipole moment $p$ in an electric field $E$. It is rotated from $\theta _{1}$ to $\theta _{2}$. We have to find the work done by external force.
When a dipole of dipole moment  $p$ is placed in electric field, work done in rotated the dipole by angle $\theta$ is
$W=-pE \cos{\theta _{1}}$
Now work done in rotating dipole by $\theta _{1}$ is 
$W _{2}=-pE\cos{\theta _{2}}$
Work done in rotating the dipole from $\theta _{1}$ to $\theta _{2}$ is
$W=W _{2}-W _{1}$
$=-pE \cos{\theta _{2}}-(-pE \cos{\theta _{1}})$
$=pE(\cos{\theta _{1}}-\cos{\theta _{2}})$
Multiple choice torque on a dipole in a uniform electric field electric dipole electric charges and fields electrostatics physics

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however not constant but increases uniformly along the positive z-direction at the rate ${10^5}\,V/m.$ The force and the torque experienced by a system having a total dipole moment equal to ${10^{ - 7}}C - m$ in the negative z-direction is given by respectively.

  1. 0.01,0

  2. 0.02,0

  3. 0,0.01

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$z$ direction positive rate $={ 10 }^{ 5 }V/m$

torque $=$ M $\times$ $E$
            $={ 10 }^{ 5 }\times { 10 }^{ -7 }$
            $=0.01cm$

Multiple choice torque on a dipole in a uniform electric field electric dipole electric charges and fields electrostatics physics

 An electric dipole consist of two opposite charges each of magnitude $1\mu C$ separated by a distance of $2\,cm.$ The dipole is placed in an external field of ${10^5}{\text{N/C}}$.The maximum torque on the dipole is:

  1. $2 \times {10^{ - 4}}J$
  2. $2 \times {10^{ - 3}}J$
  3. $4 \times {10^{ - 3}}J$
  4. ${10^{ - 3}}N\,m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
An electric dipole consist at two opposite charge each of magnitude $=1\mu C=1\times { 10 }^{ -6 }C$
distance $=2cm$
Exter field $={ 10 }^{ 5 }N/C$
maximum torque on the dipole $=?$
$q=1\times { 10 }^{ -6 }C,\quad 2a=2cm$
                                or,  $=0.02cm$
$\therefore$    $P=q\times 2a$
           $=\left( 1\times { 10 }^{ -6 } \right) \times 0.02$
           $=2\times { 10 }^{ -8 }cm$
Intensity of the external electric field, $E=1.0\times { 10 }^{ 5 }N/C$
(i) ${ Z } _{ max }=pE=\left( 2\times { 10 }^{ -8 } \right) \left( 10\times { 10 }^{ 5 } \right) =2\times { 10 }^{ -3 }N-m$
(ii) Net work done in turning the dipole from ${ 0 }^{ 0 }$ to ${ 180 }^{ 0 }$
i.e  $W=\int _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }{ \overline { r }  } d\theta =\int _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }{ pE\sin\theta  } d\theta $
           $=pE{ \left[ -cos\theta  \right]  } _{ { 0 }^{ 0 } }^{ { 180 }^{ 0 } }$
           $=-pE\left( { \cos180 }^{ 0 }-\cos{ 0 }^{ 0 } \right) $
           $=2pE$
           $=2\times \left( 2\times { 10 }^{ -8 } \right) \left( 1\times { 10 }^{ 5 } \right) J$
           $=4\times { 10 }^{ -3 }J$
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

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 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

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

Surface charge density of a disc is given by $\sigma = \dfrac{\sigma _o r}{R}$ where $\sigma _0$ is a constant, $r$ is distance from the center, and $R$ is the radius of the disc. If electric potential at the centre of the disc is $\dfrac{\sigma _0 R}{\alpha \epsilon _0}$ then find value of $\alpha$.

  1. $2$
  2. $1$
  3. $4$
  4. $0.5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Integrating the potential contribution from concentric ring elements of charge dq over the disc yields the electric potential at the center. Solving the integration with the given surface charge density results in alpha equal to 4.

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

The ratio of the forces between two small spheres with constant charges, in air and in a medium of dielectric constant $K$, is

  1. $1:K$
  2. $K:1$
  3. $1:K^{2}$
  4. $:k{2}:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electrostatic force between charges in a medium is inversely proportional to the dielectric constant K of the medium. The ratio of the force in air to the force in the medium is therefore K:1.

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

Two conducting spheres of radii $r _1$ and $r _2$ are charged to the same surface charge density. The ratio of electric fields near their surface is -

  1. $\dfrac { { r } _{ 1 }^{ 2 } }{ { r } _{ 2 }^{ 2 } } $
  2. $\dfrac { { r } _{ 2 }^{ 2 } }{ { r } _{ 1 }^{ 2 } } $
  3. $\dfrac{r _1}{r _2}$
  4. $1:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field near the surface of a charged conducting sphere is given by E = sigma / epsilon_0, where sigma is the surface charge density. Since the spheres are charged to the same surface charge density, the electric fields near their surfaces are equal, yielding a ratio of 1:1.