Question

# Derive an expression for intensity of electric field at a point broadside position or an equatorial line of an electric dipole.

Solution

## Now, suppose that the point P is situated on the right-bisector of the dipole AB at a distance r metre from its mid-point 0 (Fig. (a)] Again Let $$E_1$$ and $$E_2$$ be the magnitudes of the intensities of the electric field at P due to the charges $$+ q$$ and $$- q$$ of the dipole respectively. The distance of P from each charge is $$\sqrt{r^2+l^2}$$. Therefore,and $$E_1=\dfrac{1}{4\pi \varepsilon _0}\dfrac{q}{(r^2+l^2)}$$ away from $$+q$$The magnitudes of $$E_1$$ and $$E_2$$ are equal (but directions are different). On resolving $$E_1$$ and $$E_2$$ into two components parallel and perpendicular to AB, the components perpendicular to AB ($$E_1\,\, sin \theta \,\,and\,\, E_2 sin \theta$$) cancel each other (because they are equal and opposite), while the components parallel to AB ($$E_1\,\, cos \theta \,\, and \,\,E_2\,\, cos \theta$$ ), being in the same direction, add up [Fig. (b)]. Hence the resultant intensity of electric field at the point P is$$E=E_1 cos \theta +E_2 cos\theta$$$$=\dfrac{1}{4 \pi\varepsilon _0}\dfrac{q}{(r^2+l^2)}cos \,\theta +\dfrac{1}{4 \pi \varepsilon _0}\dfrac{q}{(r^2+l^2)}cos \theta$$$$=\dfrac{1}{4 \pi\varepsilon _0}\dfrac{q}{(r^2+l^2)}2 cos \, \theta$$But 2ql=p (moment of electric dipole)$$\therefore$$    $$=\dfrac{1}{4 \pi\varepsilon _0}\dfrac{p}{(r^2+l^2)^{32}}$$The direction of electric field E is 'antiparallel' to the dipole axis.If $$r$$ is very large compared to $$2l$$ ($$r >>2l$$), then $$l^2$$ may be neglected in comparison to $$r^2$$.$$E=\dfrac{1}{4 \pi\varepsilon _0}\dfrac{p}{r^3}$$Newton/CoulombPhysics

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