Tag: electrostatics

Questions Related to electrostatics

An electric dipole of moment $ \vec { p } $ is placed in a uniform electric field$\vec { E }$ . Then
(i) The torque on the dipole is $\vec { p } \times \vec { E } $.
(ii) The potential energy of the system is $\vec { p } \cdot \vec { E } $.
(iii) The resultant force on the dipole is zero.
Choose the correct option.

  1. (i), (ii) and (iii) are correct

  2. (i) and (iii) are correct and (ii) is wrong

  3. Only (i) is wrong

  4. (i) and (ii) are correct and (iii) is wrong


Correct Option: A
Explanation:

In a uniform electric field, $\vec { E } $, dipole experiences a torque $\vec { \tau  } $ given by $\vec { \tau  } =\vec { p } \times \vec { E } $ but experiences no force. The potential energy of the dipole in a uniform electric field $\vec { E } $ is $U=-\vec { p } .\vec { E } $

An electric dipole of length 20 cm having $\pm3 \times { 10 }^{ -3 }$ C charge placed at 60 with respect to a uniform electric field experiences a torque of magnitude 6 N m. the potential energy of the dipole is:

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

  2. 5$\sqrt { 3 }$ J

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

  4. 3$\sqrt { 5 }$ J


Correct Option: A
Explanation:

Here,

Length of dipole $2a= 20\ cm=0.2\ m$
Charge, $q= \pm 3\times 10^{-3}\ C$,             $\theta=60^o$
Torque, $\tau=6\ Nm$
As, $\tau=pEsin\ \theta$
or $E=\dfrac{\tau}{psin \theta}= \dfrac{\tau}{q(2a)sin \theta}(\because p=q(2a)$

$\therefore E=\dfrac{6}{3\times 10^{-3}\times 20 \times 10^{-2} \times sin\ 60^o}= \dfrac{10^5}{5\sqrt 3} NC^{-1}$

Potential energy of dipole, $U=-pEcos \theta=-q(2a) E cos \theta $

$\implies U= -3\times 10^{-3}(20 \times 10^{-2} )\dfrac{10^5}{5\sqrt 3} cos 60^o= -2\sqrt 3\ J$

An electric dipole is kept on the axis of a uniformly charged ring at distance $\frac{R}{\sqrt{2}}$ from the centre of the ring. The direction of the dipole moment is along the axis. The dipole moment is P, charge of the ring is Q & radius of the ring is R. The force on the dipole is ___________________.

  1. (a) $\frac{4 k P Q}{3\sqrt{3} R^{2}}$

  2. (b) $\frac{4 k P Q}{3\sqrt{3} R^{3}}$

  3. (c) $\frac{2 k P Q}{3\sqrt{3} R^{2}}$

  4. (d) zero


Correct Option: A

Two electric dipoles of dipole moment $2 \space cm$ and $4 \space cm$ respectively are kept inside a cube of side $'a' \space m$. Total electric flux linked with the cube is (in SI units)

  1. $\dfrac{6}{\varepsilon _0}$

  2. $\dfrac{2}{\varepsilon _0}$

  3. $\dfrac{4}{\varepsilon _0}$

  4. none of these


Correct Option: D
Explanation:

Since net charge $=0$

$\therefore \phi =0$
$\therefore$ option $D$ is correct answer.

An electric dipole is kept in a non-uniform electric field. It experiences

  1. a force and a torque

  2. a force but not a torque

  3. a torque but no force

  4. neither a force nor a torque


Correct Option: A
Explanation:

In a non uniform electric field, net force$\ne0$, and net torque $\ne 0$.

An electric dipole placed with its axis in the direction of a uniform electric field experiences:

  1. a force but not torque

  2. a torque but no force

  3. a force as well as a torque

  4. neither a force nor a torque


Correct Option: D
Explanation:

Total force of on diopole $= E\ q +E( -q )$ $= 0$
We know,
 Torque $= \vec{P} \times \vec{E}$
but axis of dipole is in the direction of electric field
$\therefore \tau = P.E\ sin\ 0$
$\therefore \tau = 0$
$\therefore $ No torque, No force is acting on dipole

An electric dipole of moment $p$ is lying along a uniform electric field $E$. The work done in rotating the dipole by $90^{o}$ is:

  1. $pE$

  2. $\sqrt{2}\ pE$

  3. $\dfrac{pE}{2}$

  4. $2p\ E$


Correct Option: A
Explanation:

$Work done=\Delta pE$

$pE=-\vec p\vec E$
$pE _f=0$
$pE _i=-pE$
$\Delta pE=pE$

An electric dipole of momentum $3 \times {10}^{-8}\ Cm$ is placed in an electric field of $6 \times {10}^{4}\ N/C$ with is axis making an angle of $30^o$ with the field . Find the torque acting on the dipole.

  1. $ (9 \times 10^{-5} )N-m$

  2. $ (9 \times 10^{-4} )N-m$

  3. $ (9 \times 10^{-3} )N-m$

  4. $ (90 \times 10^{-8} )N-m$


Correct Option: B
Explanation:

We know that Torque $(\vec \tau) = \vec p \times \vec E$


$  \tau = pE\sin \theta$
$ = (3\times 10^{-8})\times ( 6\times 10^4) \sin30^0$
$ = (9 \times 10^{-4} )N-m$

A neutral water molecule $ (H _2 O) $ in its vapour state has an electric dipole moment of $ 2 \times 10^{-24} C-m. $ If the molecule is placed in an electric field of $ 2 \times 10^{4} NC^{ _1} $ , the maximum torque that the field can exert on it is nearly.

  1. $ 4 \times 10^{-20} N-m $

  2. $ 6 \times 10^{-20} N-m $

  3. $ 8 \times 10^{-20} N-m $

  4. $ 2 \times 10^{-18} N-m $


Correct Option: A
Explanation:

$P = 2 \times 10^{-24}Cm$


$E = 2 \times 10^{4}NC$

$\tau = EP = 2 \times 10^{-24} \times 2 \times 10^{4} = 4 \times 10^{-20}Nm$

A dipole is placed parallel to the electric field. If W is the work done in rotating the dipole by $60$, then the work done in rotating it by $180$ is:

  1. $ 2W$

  2. $ 3W$

  3. $ 4W$

  4. $ W/2$


Correct Option: C
Explanation:

Work done in rotating dipole by $\theta \,$ angle from eqbm. ,$W = PE\left( {1 - \cos \theta } \right)$

$\begin{array}{l}W = PE\left( {1 - \cos 60^\circ } \right)\ = PE\left( {1 - \frac{1}{2}} \right) = \frac{{PE}}{2}\or\,\,\,PE = 2W\W' = PE\left( {1 - \cos 180^\circ } \right)\ = PE\left( {1 - \left( { - 1} \right)} \right)\ = 2PE = 2 \times 2W = 4W\end{array}$