Question

Class 12th !!

Derive the expression for the electric field due to dipole :

★ at a point on the axial point
★ at a point on equatorial line

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Answer 1

the above first 2 pictures shows the figure and derivation for electric field due to a dipole at an equatorial point respectively.

the third picture shows the electric field due to a dipole at an axial point respectively.

hope it helped.

thank you.

Answer 2

12th/Physics

Electrostatics

Answer :

Electric field is the region around charged particle or charged body in which if another charge is placed, it experiences electrostatic force.

Let two point charges -q and +q are placed at distance 2a apart from each other.

★ Field at a point on the axial point

$\leadsto\sf\:\vec{E}=\vec{E_+}+\vec{E_-}$

$\leadsto\sf\:\vec{E}=\dfrac{kq}{(r-a)^2}\:\hat{p}+\dfrac{kq}{(r+a)^2}\:(-\hat{p})$

$\leadsto\sf\:\vec{E}=kq{\huge(}\dfrac{1}{(r-a)^2}+\dfrac{1}{(r+a)^2}{\huge)}\:\hat{p}$

$\leadsto\sf\:\vec{E}=kq\times \dfrac{r^2+2ra+a^2-r^2+2ra-a^2}{(r^2-a^2)^2}\:\hat{p}$

$\leadsto\sf\:\vec{E}=\dfrac{kq(4ra)}{(r^2-a^2)^2}\:\hat{p}$

$\leadsto\sf\:\vec{E}=\dfrac{2k(2aq)r}{(r^2-a^2)^2}\:\hat{p}$

$\leadsto\bf\:\vec{E}=\dfrac{2kpr}{(r^2-a^2)^2}\:\hat{p}$

If r >> a → (r² - a²)² = r³

$\leadsto\:\boxed{\bf{\pink{\vec{E}=\dfrac{2kp}{r^3}\:\hat{p}}}}$

★ Field at a point on equatorial line

$\leadsto\sf\:\vec{E}=\vec{E_+cos\theta}+\vec{E_-cos\theta}$

$\leadsto\sf\:\vec{E}=-{\huge(}\dfrac{kq}{(r-a)^2}+\dfrac{kq}{(r+a)^2}{\huge)}cos\theta\:\hat{p}$

$\leadsto\sf\:\vec{E}=-{\huge(}\dfrac{2kq}{(r^2+a^2)}{\huge)}cos\theta\:\hat{p}$

$\leadsto\sf\:\vec{E}=\dfrac{-2kq}{(r^2+a^2)}\times\dfrac{a}{(r^2+a^2)^{\frac{1}{2}}}\:\hat{p}$

$\leadsto\bf\:\vec{E}=\dfrac{-kp}{(r^2+a^2)^{\frac{3}{2}}}\:\hat{p}$

If r >> a → $\sf{(r^2+a^2)^{\frac{3}{2}}}$ = r³

$\leadsto\:\boxed{\bf{\purple{\vec{E}=\dfrac{-kp}{r^3}\:\hat{p}}}}$

Remember :

• The direction of the electric field at an axial point of an electric dipole is same as that of its dipole moment and at an equatorial point it is opposite to that of dipole moment.

Cheers!