Relation Between Potential and Electric Dipole

Potential due to an electric dipole:

Let us consider charge $-q$ is placed at point $P$ and charge $+q$ is placed at point $Q

Figure:8.a

Electric potential at point $R$ due to electric dipole would be sum of potential due to both the charges $+q$ and $-q$. So,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo>=</mo><mo>&#xA0;</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfenced><mrow><mfrac><mi>q</mi><msub><mi>r</mi><mn>1</mn></msub></mfrac><mo>-</mo><mfrac><mi>q</mi><msub><mi>r</mi><mn>2</mn></msub></mfrac></mrow></mfenced></math> ………………….. (1)

Draw two perpendiculars $PC$ and $QD$from point $P$ and $Q$ as shown.

From triangle $POC$
$\cos \theta =OC/OP=OC/a$

Or, $OC=a cos\theta $

Similarly, $OD=a cos\theta $
Now,

$r_{1}=QR\approx RD=OR-OD=r-a cos\theta $

And

$r_{2}=PR\approx RC=OR+OC=r+a cos\theta $

From equation (1), we get,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo>=</mo><mo>&#xA0;</mo><mfrac><mi>q</mi><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfenced><mrow><mfrac><mn>1</mn><mrow><mi>r</mi><mo>+</mo><mi>a</mi><mo>&#xA0;</mo><mi>cos</mi><mi>&#x3B8;</mi></mrow></mfrac><mo>-</mo><mfrac><mn>1</mn><mrow><mi>r</mi><mo>-</mo><mi>a</mi><mo>&#xA0;</mo><mi>cos</mi><mi>&#x3B8;</mi></mrow></mfrac></mrow></mfenced><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mfrac><mi>q</mi><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfenced><mfrac><mrow><mn>2</mn><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>cos</mi><mi>&#x3B8;</mi></mrow><mrow><msup><mi>r</mi><mn>2</mn></msup><mo>-</mo><msup><mi>a</mi><mn>2</mn></msup><msup><mi>cos</mi><mn>2</mn></msup><mi>&#x3B8;</mi></mrow></mfrac></mfenced><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mfrac><mi>q</mi><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><mn>2</mn><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>cos</mi><mi>&#x3B8;</mi></mrow><msup><mi>r</mi><mn>2</mn></msup></mfrac></math>

(Consider $r>>a$)

The magnitude of dipole is,$\vert p\vert =2qa$

So, the above equation becomes,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo>=</mo><mo>&#xA0;</mo><mfrac><mrow><mi>p</mi><mo>&#xA0;</mo><mi mathvariant="bold">cos</mi><mi mathvariant="bold-italic">&#x3B8;</mi></mrow><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub><msup><mi>r</mi><mn>2</mn></msup></mrow></mfrac><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mfrac><mrow><mi>p</mi><mo>.</mo><mover><mi>r</mi><mo>^</mo></mover></mrow><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub><msup><mi>r</mi><mn>2</mn></msup></mrow></mfrac></math>

Where ${r}$ is the unit vector along the vector ${OR}$.
From the above equation, we can see that the electric potential due to electric dipole does not only depend on $r$ but also depends on the angle between position vector $r$ and dipole moment $p$.

Work done in rotating an electric dipole in an electric field:

Consider a dipole placed in a uniform electric field. If the dipole is rotated from its equilibrium position, work has to be done.

Figure:8.b

Let the dipole of the moment $p$ is rotated through an angle $\theta $ from its equilibrium position.

The torque acting on the dipole is,

$\tau =pE sin\theta $

Work done when the dipole makes a small angle $d\theta $ is given by,

$dW=pE sin\theta d\theta $
Total work done is,$W=_{0}^{\theta }{}pE sin\theta d\theta $

$W=pE\lbrack 1-cos\theta \rbrack $

This is the expression for work done in rotating an electric dipole placed in a uniform electric field $E$ through an angle $\theta $ from its equilibrium.

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