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

Question

Vedantu Content Team · Accepted Answer

Hint: 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.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:\n    \n    \n    \n    \n  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 BPAlso, 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.

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

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