Two short electric dipoles A and B having dip | vedclass

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(B) For a short electric dipole, the electric field at a point at distance $r$ and angle $	heta$ from the dipole axis is given by $\vec{E} = E_r \hat

Vedclass · Accepted Answer

$2\sqrt{3}$

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(B) For a short electric dipole, the electric field at a point at distance $r$ and angle $	heta$ from the dipole axis is given by $\vec{E} = E_r \hat{r} + E_	heta \hat{	heta}$, where $E_r = \frac{2kp\cos	heta}{r^3}$ and $E_	heta = \frac{kp\sin	heta}{r^3}$.For dipole $A$, the point $x$ lies on its axial line, so $	heta = 0^{\circ}$. The electric field due to $A$ is $E_A = \frac{2kp_1}{r^3}$ directed along the line $Ox$.For dipole $B$, the point $x$ lies on its equatorial line, so $	heta = 90^{\circ}$. The electric field due to $B$ is $E_B = \frac{kp_2}{r^3}$ directed perpendicular to the line $Ox$.The resultant electric field makes an angle of $60^{\circ}$ with the line $Ox$. Therefore, $	an 60^{\circ} = \frac{E_B}{E_A}$.Substituting the values, we get $\sqrt{3} = \frac{kp_2/r^3}{2kp_1/r^3} = \frac{p_2}{2p_1}$.Thus, the ratio $\frac{p_2}{p_1} = 2\sqrt{3}$.

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