Electric Dipoles in Electric Fields - Torque, Work and Energy

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 } $

Here,

The electric field on the axis of a ring is E = kQx / (R^2 + x^2)^(3/2). For a dipole at x = R/sqrt(2), the field gradient exerts a force F = p * (dE/dx). Calculating the derivative and substituting gives the force.

Since net charge $=0$

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

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

$Work done=\Delta pE$

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

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

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

Torque is given by p x E. First, calculate the dipole moment vector p = q * d, where d is the vector from -q to +q. Then compute the cross product with the electric field E.

Work done in rotating an electric dipole from angle theta_1 to theta_2 in a uniform electric field is W = pE(cos(theta_1) - cos(theta_2)). Initially, it is parallel (theta = 0) and rotated by 60 degrees, so W = pE(cos(0) - cos(60)) = pE(1 - 0.5) = 0.5 pE. Rotating it by 180 degrees gives W_180 = pE(cos(0) - cos(180)) = pE(1 - (-1)) = 2 pE. Since 2 pE = 4 * (0.5 pE), the work done is 4 W.

$\tau =\vec P\times \vec E = PE \sin\theta $
Since, $ \theta = 180^o \rightarrow \tau =0 $

Work done due to dipole $=W=-pE\cos\theta$ where $p=$ dipole moment, $E=$ electric field
Here, $W=W _{(180)}-W _{(0)}=(-pE\cos180)-(-pE\cos0)=pE+pE=2pE$
Ans: (B)

Work done $=PE\cos{ 60 }^{ 0 }$

In a uniform electric field, the net force on a dipole is zero, so it only experiences rotation. In a non-uniform field, the forces on the two charges are unequal, leading to both a net force (linear displacement) and a torque (rotational motion).

Maximum torque is given by tau = pE, where p = q * d. Here, p = (1.0 * 10^-6 C) * (0.02 m) = 2.0 * 10^-8 C-m. Then, tau = (2.0 * 10^-8) * (1.0 * 10^5) = 2.0 * 10^-3 N-m.

Dipole means two charges $+q$ and $-q$ are placed at a certain distance.
There  three cases may be occured for electric field whose direction is fixed but magnitude varies with distance.
1) when forces on the two charges are in different direction and different magnitude, they will have a net force and a torque.
2) when forces on the two charges are in same direction but different magnitude, they will have a net force but no torque.
3) when forces on the two charges are in opposite direction but same magnitude, they will have no net force but a torque.

Electric field will produce a torque on the dipole.
Torque , $\displaystyle \tau=pE\sin\theta$
The work done in rotating it from an equilibrium position by an angle $\theta$ is $\displaystyle W=\int _0^{\theta}\tau d\theta=pE\int _0^{\theta}\sin\theta d\theta=-pE(\cos\theta-\cos0)=pE(1-\cos\theta)$

The electric field produced by a dipole is non-uniform. Therefore, another dipole placed in this field will experience a non-zero net force and a torque, depending on its orientation and position.

Torque is given by tau = pE sin(theta). When the dipole is at 180 degrees to the field, sin(180) = 0, resulting in zero torque.

Maximum torque is tau = pE. Given q = 10^-6 C, d = 0.02 m, E = 10^5 N/C. p = qd = 2 * 10^-8 C-m. tau = (2 * 10^-8) * 10^5 = 2 * 10^-3 N-m.

Dipole moment of the dipole p and uniform Electric field $E$. we know that dipole moment $p = qa$ (where q is charge and a is the dipole length).And when a dipole of Dipole moment p is placed in a uniform Electric field E , the torque $\tau = Either  force \times \text{perpendicular  distance  between  the  two  forces }= qaE sin \theta$ or $\tau = pEsin \theta$ or $\tau = p \times E$

The work done in rotating a dipole from theta_1 to theta_2 is W = pE(cos(theta_1) - cos(theta_2)). For a rotation from 0 to 60 degrees, W = pE(1 - 0.5) = 0.5 pE. For a rotation from 0 to 180 degrees, W_total = pE(1 - (-1)) = 2 pE. Comparing the two results, 2 pE is exactly four times 0.5 pE, making the work 4 W.

A and B cannot be true in any case. Because for the net force on a dipole to be zero, the field lines should be parallel to each other and the dipole must be placed perpendicular to the field. However, in case of point charge you cannot have parallel field lines. 
C: Not true again. Because in any orientation (except radial) the force on the positive and negative charge will form a couple and you will have torque.
D: True. If you place the dipole radial from the point object, then there will be no net torque. Since torque $=F\times R;\sin{\theta}$. If it is radial, $\sin{\theta}=0$, so torque $=0$. 

Given an electric dipole placed in a non-uniform electric field. An electric dipole always experiences a torque when placed in uniform as well as non-uniform electric field. But in non-uniform electric field, dipole will also experience net force of attraction. So the electric dipole in non-uniform electric field experiences both torque and force.

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

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

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

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.

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

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

Work done in rotating a dipole in a uniform field is W = integral of tau d(theta) from 0 to theta, which equals pE(1 - cos(theta)).

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

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

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

A dipole in a point charge field experiences a torque that is zero only when the dipole axis is aligned with the radial line from the point charge. Thus, it is not always zero, but may be zero.

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