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Potential energy of a dipole in external field - class-XII

Description: potential energy of a dipole in external field
Number of Questions: 20
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Tags: electrostatics physics electric charges and fields
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An electric dipole of length $20cm$ having $\pm 3\times { 10 }^{ -3 }C$ charge placed at ${60}^{o}$ with respect to a uniform electric field experiences a torque of magnitude $6Nm$. The potential energy of the dipole is

  1. $-2\sqrt{3}J$

  2. $5\sqrt{3}J$

  3. $-2\sqrt {2}J$

  4. $3\sqrt {5}$


Correct Option: A
Explanation:

Here length of dipole $2a=20cm=20\times { 10 }^{ -2 }m$, Charge $q=\pm 3\times { 10 }^{ -3 }C,\theta ={ 60 }^{ o }\quad $ and torque $\tau =6Nm$
As $\tau =pE\sin { \theta  } $
or $E=\cfrac { \tau  }{ p\sin { \theta  }  } =\cfrac { \tau  }{ q(2a)\sin { \theta  }  } \left( \because p=q(2a) \right) $
$\therefore E=\cfrac { 6 }{ 3\times { 10 }^{ -3 }\times 20\times { 10 }^{ -2 }\times \sin { { 60 }^{ o } }  } =\cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 }  } N{ C }^{ -1 }$
Potential energy of dipole $U=-pE\cos{\theta}=-q(2a)E\cos{\theta}$
$=-3\times { 10 }^{ -3 }\left( 20\times { 10 }^{ -2 } \right) \cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 } } \cos { { 60 }^{ o } } =\cfrac { -3\times { 10 }^{ -5 }\times 20\times { 10 }^{ 5 } }{ 5\sqrt { 3 } \times 2 } =-2\sqrt { 3 } J\quad \quad $

An electric dipole has the magnitude of its charge as $q$ and its dipole moment is $p$. It is placed in uniform electric field $E$. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively

  1. $q.E$ and max

  2. $2q.E$ and min.

  3. $q.E$ and min

  4. zero and min.


Correct Option: D
Explanation:

When the dipole is in the direction of field then net force is $qE+(-qE)=0$
and its potential energy is minimum $=-p.E$
$=-qaE$

An electric dipole of diploe moment $\overrightarrow { p } $ placed in uniform electric field $\overrightarrow { E } $ has minimum potential energy when angle between $\overrightarrow { p } $ and $\overrightarrow { E } $

  1. $\cfrac{\pi}{2}$

  2. zero

  3. $\pi$

  4. $\cfrac{3\pi}{2}$


Correct Option: C
Explanation:

Potential Energy=$ -PE \cos {\theta}$

when 
$ \theta=0 $
Potential Energy=$ -PE $
When
$ \theta=180 $
Potential Energy=$ +PE $
So, Maximum Potential Energy=$ +PE $ at angle $\theta=\pi$

 Two small electric dipoles each of dipole moment pi are situated at $(0, 0, 0)$ and $(r, 0, 0)$. the electric potential at a point $\left( \frac { r } { 2 } , \frac { \sqrt { 3 } r } { 2 } , 0 \right)$ is:

  1. $\frac { p } { 4 \pi \in _ { 0 } r ^ { 2 } }$

  2. $0$

  3. $\frac { p } { 2 \pi \epsilon _ { 0 } r ^ { 2 } }$

  4. $\frac { p } { 8 \pi \epsilon _ { 0 } r ^ { 2 } }$


Correct Option: D

Potential at any point in the electric field produced by a dipole is

  1. $\infty , r$

  2. $\alpha r ^ { 2 }$

  3. $\frac { 1 } { r }$

  4. $\frac { 1 } { r ^ { 2 } }$


Correct Option: A

A dipole of dipole moment $\overline {\text{p}} $ i s aligned at right angle to electrictric field $\overline {\text{E}} $ . To set it at an angle $\theta $ with E the amount of work done is


  1. $ - {\text{pEcos}}\theta $

  2. $ {\text{pEsin}}\theta $

  3. $ - {\text{pE}}\left( {{\text{sin}}\theta - 1} \right)$

  4. $ - {\text{pE}}\left( {{\text{sin}}\theta + 1} \right)$


Correct Option: A

A electric dipole moment $\vec { p } =\left( 2.0\hat { i } +3.0\hat { j }  \right) \mu C.m$ is placed in a uniform electric field $\vec { E } =\left( 3.0\hat { i } +2.0\hat { k }  \right) \times { 10 }^{ 5 }N{ C }^{ -1 }$

  1. The torque that $\vec { E }$ exerts on $\vec { p }$ is $\left( 0.6\hat { i } -0.4\hat { j } -0.9\hat { k } \right) Nm$

  2. The potential energy of the dipole is $-0.6J$

  3. The potential energy of the dipole is $0.6J$

  4. If the dipole is free to rotate in the electric field, the maximum magnitude of potential energy of the dipole during the rotation is $1.3J$


Correct Option: A,B,D
Explanation:

$\vec P = (2 \widehat i + 3 \widehat j) \mu cm$.
$\vec E = (0.3 \widehat i + 0.2 \widehat k) N \mu C^{-1}$
$\vec C = \vec P \times \vec E$
$=\begin{vmatrix}\widehat i & \widehat j & \widehat k\ 2 & 3 & 0\0.3  & 0 & 0.2\end{vmatrix}$
$= \widehat i (0.6) - \widehat (0.4) + \widehat k (-0.9)$
$= 0.6 \widehat i - 0.4 \widehat j - 0.9 \widehat k$
$U =- \vec P \cdot \vec E$
$=- (2 \widehat i + 3 \widehat j) \cdot (0.3 \widehat i + 0.2 \widehat k)$
$=- 0.6 J$
Consider,
the dipole rotated by 180$^o$.

The magnitude of dipole moment will not change only its direction will change.
$\therefore U=-\vec{P}.\vec{E}=|P||E|\ Sin\theta$
the max value of U is $|P||E|$
$=\sqrt{13} \times 10^{-6} \times \sqrt{13} \times 10^5
= 1.3 J.$
This is the maximum potential energy of the dipole.

 An electric dipole of moment $P$ is placed in the position of stable equilibrium in uniform electric field of intensity $E$. It is rotated through an angle $\theta$ from the initial position. The potential energy of electric dipole in the position is

  1. $\mathrm { pE } \cos \theta$

  2. $\mathrm { pE } \sin \theta$

  3. $\mathrm { pE } ( 1 - \cos \theta )$

  4. $\mathrm {- pE } \cos \theta$


Correct Option: C

 A small dipole is placed is located at the center of an imaginary spherical Gaussian surface (radius R) with its dipole moment in +X-direction . Let $E _{max}$ & $E _{min}$  be maximum & maximum possible magnitude of field over the surface. 
Statement 1:   Number of points where E = $E _{max}$ is infinite.
Statement 2:    Number of points where E = $E _{min}$ is two.

  1. Both 1 and 2 are correct

  2. Both 1 and 2 are incorrect

  3. Only 1 is correct

  4. Only 2 is correct


Correct Option: C

If $ P= 2 \times 10^7 cm $ of an electric dipole placed in an uniform electric field of intensity $ 1 \times  10^8 N/C $ making an angle $ 60^0 $  with electric field. find magnitude of potential energy____J?

  1. $ 10^{-3} $

  2. $ 10^{-4} $

  3. $ 1.73 \times 10^{13} $

  4. $ 10^{2} $


Correct Option: C

An electric dipole consists of two opposite charges each of magnitude $2\mu C$ separated by a distance $1cm$. The dipole is placed in an external field of $10^3N/C$. The maximum torque on the dipole is

  1. $1\times 10^{-5}N-m$

  2. $2\times 10^{-5}N-m$

  3. $0.5\times 10^{-5}N-m$

  4. $Zero$


Correct Option: B

The relation connecting the energy U and distance r between dipole and induced dipole is :

  1. $U\propto r$

  2. $U\propto r^{2}$

  3. $U\propto r^{-6}$

  4. $U\propto r^{6}$


Correct Option: C
Explanation:

The potential energy for the dipole-dipole interaction is given by $\displaystyle U=-\dfrac{2p _1^2p _2^2}{3(4\pi\epsilon _0)^2k _BT r^6}$
thus, $U \propto r^{-6}$

An electric dipole moment $ \overrightarrow { P }  $ is lying a uniform electric field $ \overrightarrow { E }  $ .The work done in rotation the dipole by $ 37^o $

  1. $ \dfrac {2}{5} PE $

  2. $ - \dfrac {2}{5} PE $

  3. $ \dfrac {PE}{5} $

  4. $ \dfrac {3}{5} PE $


Correct Option: C

An electric dipole is placed in an electric field generated by a point charge then

  1. Then net electric force on the dipole must be zero

  2. The net electric force on the dipole may be zero

  3. The torque on the dipole due to the field may be zero

  4. Both (2) and (3)


Correct Option: C

An electric dipole when placed in a uniform electric field $E$ will have a minimum potential energy if the dipole moment makes the following angle with $E$

  1. $\pi$

  2. $\pi /2$

  3. zero

  4. $3\pi /2$


Correct Option: C
Explanation:

${ U } _{ p }=-p\bullet E=-pE\cos { \theta  } $
${ \left( { U } _{ p } \right)  } _{ minimum }=-pE$
$\theta ={ 0 }^{ o }$${ U } _{ p }=-p\bullet E=-pE\cos { \theta  } $
${ \left( { U } _{ p } \right)  } _{ minimum }=-pE$
$\theta ={ 0 }^{ o }$

An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively:

  1. q. E and p. E

  2. zero and minimum

  3. q. E and maximum

  4. 2q. E and minimum


Correct Option: B
Explanation:

$F = p\dfrac{dE} {dr} = 0 \left ( \because E = constant \right )$
$u = -\overrightarrow{p} \overrightarrow{E} = -PE \left ( minimum \right )$

Intensity of an electric field (E) depends on distance $r$. In case of dipole, it is related as :

  1. $ E \propto \cfrac{1}{r}$

  2. $ E \propto \cfrac{1}{r^{2}}$

  3. $ E \propto \cfrac{1}{r^{3}}$

  4. $ E \propto \cfrac{1}{r^{4}}$


Correct Option: C
Explanation:

Intensity of electric field due to a Dipole
$ E = \cfrac{p}{4\pi \varepsilon _{0}r^{3}} \sqrt{3cos^{2 }\theta+1}\Rightarrow E \propto \cfrac{1}{r^{3}}$

So, we can just dimensionally tell that Electric field will be inversely proportional to third power of $r$.

A point charge $Q$ lies on the perpendicular bisector of an electric dipole of dipole $p$. If the distance of $Q$ from the dipole is $r$ (much larger than the size of the dipole).then the electric field at $\theta$ is proportional to :

  1. $P^{2}$ and $r^{-3}$

  2. $P$ and $r^{-2}$

  3. $P^{-1}$ and $r^{-2}$

  4. $P$ and $r^{-3}$


Correct Option: D
Explanation:

$\begin{array}{l} As\, \, we\, \, have, \ if\, \, r>1 \ { P _{ axi } }=\frac { 1 }{ { 4\pi { E _{ 0 } } } } \frac { { 2P } }{ { { r^{ 3 } } } }  \ { V _{ axi } }=\frac { 1 }{ { 4\pi { E _{ 0 } } } } \frac { P }{ { { r^{ 2 } } } }  \ Where\, \, in, \ Angle\, \, between\, \, { P _{ axi } }\, \, and\, \, P\, \, is\, 0. \ { E _{ equatorial } }=\frac { { kp } }{ { { r^{ 3 } } } }  \ i.e\, \, \, E\propto p \ and\, \, P\propto { r^{ -3 } } \end{array}$

Hence, Option $D$ is correct answer.

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