Question

Derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

Answer 1

To derive the expression for electric field due to an electric dipole, consider AB is an electric dipole of two point charges – q and + q separated by small distance 2d and then consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.

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Formula used: $E = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{q}{{{x^2}}}$

Complete step-by-step answer:

We have been asked to derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

Now, let AB be an electric dipole of two-point charges -q and + q separated by a small distance 2d.

Also, consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.

Now, for better understanding refer the figure below:

The electric field at any point due to a charge q at a distance x is given by-

$E = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{q}{{{x^2}}}$

Now, the electric field at the point P due to + q charge placed at B is given by-

${E_1} = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{q}{{{{(r - d)}^2}}}$ (along BP)

Here, (r – d) is the distance of point P from charge +q.

Also, the electric field at point due to – q charge placed at A-

${E_2} = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{q}{{{{(r + d)}^2}}}$ (along PA)

Therefore, the magnitude of the resultant electric field (E) acts in the direction of the vector with a greater magnitude.

The resultant electric field at P is-

$E = {E_1} + ( - {E_2})$

Putting the values; we get-

$E = \left[ {\dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{q}{{{{(r - d)}^2}}} - \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{q}{{{{(r + d)}^2}}}} \right]$ (along BP)

Simplifying further we get-

$E = \dfrac{q}{{4\pi { \in _ \circ }}}.\left[ {\dfrac{1}{{{{(r - d)}^2}}} - \dfrac{1}{{{{(r + d)}^2}}}} \right]$

$E = \dfrac{q}{{4\pi { \in _ \circ }}}.\left[ {\dfrac{{{{(r + d)}^2} - {{(r - d)}^2}}}{{{{(r - d)}^2}{{(r + d)}^2}}}} \right]$

Now, we can write ${(r - d)^2}{(r + d)^2} = {({r^2} - {d^2})^2}$

So, we get-

$E = \dfrac{q}{{4\pi { \in _ \circ }}}.\left[{\dfrac{{{{(r)}^2} + {{(d)}^2} + 2rd - {{(r)}^2} - {{(d)}^2} + 2rd}}{{{{(r - d)}^2}{{(r + d)}^2}}}} \right]$

$E = \dfrac{q}{{4\pi { \in _ \circ }}}.\left[ {\dfrac{{4rd}}{{{{({r^2} - {d^2})}^2}}}} \right]$

Now, if the point P is far away from the dipole, then $d \ll r$

So, the electric field will be-

$E = \dfrac{q}{{4\pi { \in _ \circ }}}.\left[ {\dfrac{{4rd}}{{{{({r^2})}^2}}}} \right] = \dfrac{q}{{4\pi { \in _ \circ }}}.\dfrac{{4rd}}{{{r^4}}} = \dfrac{q}{{4\pi { \in _ \circ }}}.\dfrac{{4d}}{{{r^3}}}$

along BP

Also, electric dipole moment $p = q \times 2d$ , so the expression will now become-

$E = \dfrac{q}{{4\pi { \in _ \circ }}}.\dfrac{{2(2d)}}{{{r^3}}} = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{{2p}}{{{r^3}}}$

E acts in the direction of the dipole moment.

**Therefore, the expression for the electric field intensity due to an electric dipole at a point on its axial line is $E = \dfrac{q}{{4\pi { \in _ \circ }}}.\dfrac{{2(2d)}}{{{r^3}}} = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{{2p}}{{{r^3}}}$ **

Note:

Whenever it is required to derive an expression, then always first draw a rough sketch depicting a dipole and a point on the axial line. As mentioned in the figure, first we found out the electric field at the point P due to + q charge placed at B and then due to charge – q placed at A. Then, the resultant electric field at P is found out. After making necessary assumptions, we got the expression for electric field intensity.