Tag: electric charges and fields

Questions Related to electric charges and fields

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. ,$W = PE\left( {1 - \cos \theta } \right)$

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

Two charges of charge $-4\mu C$ and $+4\mu C$ are placed of the point $A(1, 0, 4)$ and $B(2, -1, 5)$ located in an electric field $E = 0.20\hat {i} V/ C-m$. Then, torque acting on the dipole will be

  1. $2.31\times 10^{-4} N/m$

  2. $7.98\times 10^{-4} C/m$

  3. $7.11\times 10^{-4} C/m$

  4. $7.04\times 10^{-4} C/m$


Correct Option: D