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

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 $

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

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

Work done in rotating a dipole in an electric field is W = U_final - U_initial. U = -pE cos(theta). Initial angle is 90 degrees (cos 90 = 0). Final angle is theta. W = -pE cos(theta) - 0 = -pE cos(theta).

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

Stable equilibrium is at theta = 0. Potential energy U = -pE cos(theta). Rotating to angle theta gives U(theta) = -pE cos(theta). The change in potential energy relative to the stable position is U(theta) - U(0) = -pE cos(theta) - (-pE) = pE(1 - cos(theta)).

Torque tau = pE sin(theta). Max torque occurs at theta = 90 degrees, so tau_max = pE. p = q * d = (210^-6 C) * (0.01 m) = 210^-8 C-m. E = 10^3 N/C. tau_max = (210^-8) * (10^3) = 210^-5 N-m.

Work done W = pE(cos(theta1) - cos(theta2)). Assuming rotation from 0 to 37 degrees, W = pE(cos(0) - cos(37)) = pE(1 - 4/5) = pE/5.

In a non-uniform field (like that of a point charge), the net force on a dipole is generally non-zero. However, the torque can be zero if the dipole is aligned with the radial field line.

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

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