Tag: electrostatics

Questions Related to electrostatics

The torque acting on a dipole of momentum $\vec { p } $ in an electric field $\vec { E } $:

  1. $\vec { p } \times \vec { E } $

  2. $\vec { p } .\vec { E } $

  3. zero

  4. $\vec { E } \times \vec { p } $


Correct Option: A
Explanation:

Force acting on dipole due to electric field is given by,

$F=qE$
Torque on a dipole in an electric field is given by,
$T= F\times d = Fdsin\theta=qEdsin\theta$.     .....(i)
But dipole moment is given by,
$P=qd$    ....(ii)
By (i) and (ii) we get,
$\therefore$ $T= pEsin\theta=\overrightarrow{p}\times \overrightarrow{E}$

An electric dipole of moment $\vec { p } $ is placed normal to the lines of force of electric intensity $\vec { E } $, then work done in deflecting it through an angle of ${180}^{o}$ is:

  1. $pE$

  2. $+2pE$

  3. $-2pE$

  4. Zero


Correct Option: A
Explanation:

Work done $W=pE(\cos\phi _1-\cos\phi _2)$

$\phi _1=90^\circ,\phi=180^\circ$
$W=pE(0-(-1))$
$W=pE$

An electric dipole of dipole moment $\vec { P } $ is placed parallel to the uniform electric field of intensity $\vec { E }$. On rotating it through ${180}^{o}$, the amount of work done is ________ .

  1. $2PE$

  2. Zero

  3. $PE$

  4. $-2PE$


Correct Option: A
Explanation:

Work done by external agent in rotating the dipole 
$W=PE\left[ \cos { { \theta  } _{ 1 } } -\cos { { \theta  } _{ 2 } }  \right] $

Consider the initial angle to be
${ \theta  } _{ 1 }=0$ 

and the final angle will be
 ${ \theta  } _{ 2 }={ 180 }^{ o }$

So,$\Rightarrow$ $W=PE[cos\ 0^o-cos\ 180^o]=2PE$

An electric dipole kept in a uniform electric field experiences :

  1. Force and a torque

  2. A force, but no torque

  3. A torque but no force

  4. Neither a force nor a torque


Correct Option: C
Explanation:

In a uniform electric field, both the point charges comprising the dipole will experience force, equal in magnitude and opposite in direction.
Though the net force will always be zero, the torque will be in same direction for both the charges. Hence torque will not be zero.

When an electric dipole $\vec p$ is kept in a uniform electric field $\vec E$ then for what of a value of the angle between $\vec p$ and $\vec E$, torque will be maximum:

  1. ${90^o}$

  2. ${0^o}$

  3. ${180^o}$

  4. ${45^o}$


Correct Option: A
Explanation:

$\overline { P } \times \overline { E } $

$=\overline { P } \overline { E } \sin{ 90 }^{ 0 }$
for maximum $\overline { P } \overline { E } $ then $\theta $ will be ${ 90 }^{ 0 }$.

What will be the magnitude of torque on an electric dipole having dipole moment of $4\times { 10 }^{ -9 }cm$ placed in a uniform electric field of intensity of $5\times { 10 }^{ 4 \,\,}N { C }^{ -1 }$ making an angle ${180}^{o}$ with the field.

  1. ${ 10 }^{ -4 }N-m$

  2. $2\times { 10 }^{ -4 }N-m$

  3. $0$ (zero)

  4. ${ 10 }^{ -6 }N-m$


Correct Option: C
Explanation:

$\overrightarrow { \tau  } =\overrightarrow { p } \times \overrightarrow { E } $

$\overrightarrow{\tau} \longrightarrow$torque
$[\theta \longrightarrow$angle of the dipole moment$(p)$ with the field$(E)]$
$\therefore |\tau |=pE\sin { \theta  } =pE\sin { 180° } =0$

An electric dipole  of dipole moment $\vec {p}$ is placed in uniform electric field $\vec {E}$, with $\vec {p}$ parallel to $\vec {E}$ . It is then rotated by an angle of $\theta$. The work done is

  1. $pE\ \sin \theta$

  2. $pE\ \cos \theta$

  3. $pE\ (1-\cos \theta)$

  4. $pE\ (1-\sin \theta)$


Correct Option: C

A dipole of $2 \mu C$ charges each other consists of the positive charge at the point $P(1, -1)$ and the m=negative charge is placed at the point $Q(-1,1)$ . The work done in displacing a charge of $ + 1 \mu C$ from point $A (-3,-3) $ to $B(4,4) $ is :

  1. $1.6 \times 10^{-19} J $

  2. $ 6.98 \times 10^{-3} J $

  3. Zero

  4. $4.8 eV $


Correct Option: B
Explanation:

Since origin is the central point of the dipole, so electric potential on point ${P _1}\left( { - 3,3} \right)$

${V _1} = \dfrac{{k \times P\cos \theta }}{{{r^2}}}$

${V _1} = \dfrac{{9 \times {{10}^9} \times 2\sqrt 2  \times 2 \times {{10}^{ - 6}}}}{{{{\left( {3\sqrt 2 } \right)}^2}}} \times \dfrac{{\sqrt 2 }}{{\left( {3\sqrt 2 } \right)}}$

Similarly electric potential at point ${P _2}\left( {4,\;4} \right)$

${V _2} = \dfrac{{k \times P\cos \theta }}{{{r^2}}}$

${V _2} = \dfrac{{9 \times {{10}^9} \times 2\sqrt 2  \times 2 \times {{10}^{ - 6}}}}{{{{\left( {4\sqrt 2 } \right)}^2}}} \times \dfrac{{\sqrt 2 }}{{\left( {4\sqrt 2 } \right)}}$

Change in potential

$\Delta V = {V _1} - {V _2}$

$\Delta V = 9 \times {10^9} \times 2\sqrt 2  \times 2 \times {10^{ - 6}}\left( {\dfrac{1}{{18 \times 3}} - \dfrac{1}{{4 \times 32}}} \right)$

$\Delta V = 36\sqrt 2  \times {10^3}\left( {18.52 \times {{10}^{ - 3}} - 7.81 \times {{10}^{ - 3}}} \right)$

$\Delta V = 6.98 \times {10^3}{\rm{V}}$

Since potential at point ${P _1}$ is higher than potential at point ${P _2}$. The charge will move automatically from point ${P _1}$ to point ${P _2}$ under the effect of electric field of the dipole.

Now work done

$W = 6.98 \times {10^3} \times 1 \times {10^{ - 6}}$

$W = 6.98 \times {10^{ - 3}}{\rm{J}}$

Two electric dipoles of moment $\rho $ and $64\rho $ are placed in opposite direction on a line at a distance of $25\ cm$. The electric field will be zero at point between the dipoles whose distance from dipole of moment $\rho $ is 

  1. $5\ cm$

  2. $\dfrac{25}{9}$ cm

  3. 10 cm

  4. $\dfrac{4}{13}$ cm


Correct Option: C

An electric dipole of moment 'p' is placed in an electric field of intensity 'E'. The dipole acquires a position such that the axis of the dipole makes an angle $\theta $ with the direction of the field. Assuming that the potential energy of the dipole to be zero when=${ 90 }^{ 0 }$, the torque and the potential energy of the dipole will respectively be

  1. $p E \sin { \theta } , pE\cos { \theta } $

  2. $p E \sin { \theta } ,-2p E\cos { \theta } $

  3. $p E \sin { \theta } ,2p E\cos { \theta } $

  4. $p E \cos { \theta } ,-p E\cos { \theta } $


Correct Option: A
Explanation:
An electric dipole of moment $=P$
electric field of intensity $=E$
the dipole acquire a position angle $=\theta$
dipole to be zero when $={ 90 }^{ 0 }$
torque and the potential energy of the dipole will respectively 
potential energy $U=PE\sin\theta $ in this situation.
$SIn$ component is benefited for that, perpendicularity and dipole is hence, $PEcos\theta $