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Question 90 of 122

Q.The magnitude of electric intensity at a distance 'r' from the centre of an electric dipole along its axial line is E. The distance of the point from the centre of the electric dipole along its equatorial line at which the electric intensity has the same value E is :

(a) r2\dfrac{r}{\sqrt{2}}
(b) r(3)1/3\dfrac{r}{(3)^{1/3}}
(c) r(2)1/3r(2)^{1/3}
(d) r(2)1/3\dfrac{r}{(2)^{1/3}}
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2019MCQ· 1mImportance★★★★★
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Equating the axial-line field at distance rr to the equatorial-line field gives the equatorial distance as r/21/3r/2^{1/3}.

For a short electric dipole of dipole moment pp, the electric field intensity at a point on the axial line at distance rr from the centre (for rr much greater than the dipole length) is Eaxial=14πϵ02pr3=2kpr3E_{axial} = \dfrac{1}{4\pi\epsilon_0}\dfrac{2p}{r^3} = \dfrac{2kp}{r^3}, where k=14πϵ0k = \dfrac{1}{4\pi\epsilon_0}.

The field intensity at a point on the equatorial (perpendicular bisector) line at distance r′r' from the centre is Eeq=14πϵ0pr′3=kpr′3E_{eq} = \dfrac{1}{4\pi\epsilon_0}\dfrac{p}{r'^3} = \dfrac{kp}{r'^3}.

We are told these two fields are equal in magnitude: 2kpr3=kpr′3\dfrac{2kp}{r^3} = \dfrac{kp}{r'^3}

Cancelling kpkp from both sides: 2r3=1r′3  ⟹  r′3=r32\dfrac{2}{r^3} = \dfrac{1}{r'^3} \implies r'^3 = \dfrac{r^3}{2}

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