Two point dipoles of dipole moments vec{p}_1 | vedclass

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Question

(C) The electric field produced by a dipole with moment $\vec{p}_1$ at a distance $x$ in the equatorial position is given by $\vec{E} = -\frac{1}{4\pi

Vedclass · Accepted Answer

$\frac{1}{4\pi\varepsilon_0} \frac{6p_1p_2}{x^4}$

Vedclass · Answer

(C) The electric field produced by a dipole with moment $\vec{p}_1$ at a distance $x$ in the equatorial position is given by $\vec{E} = -\frac{1}{4\pi\varepsilon_0} \frac{\vec{p}_1}{x^3}$.The potential energy of the second dipole $\vec{p}_2$ in this field is $U = -\vec{p}_2 \cdot \vec{E} = -\vec{p}_2 \cdot \left( -\frac{1}{4\pi\varepsilon_0} \frac{\vec{p}_1}{x^3} \right) = \frac{1}{4\pi\varepsilon_0} \frac{p_1p_2}{x^3}$ (since $\vec{p}_1 \parallel \vec{p}_2$).The force between the dipoles is given by $F = -\frac{dU}{dx}$.$F = -\frac{d}{dx} \left( \frac{1}{4\pi\varepsilon_0} \frac{p_1p_2}{x^3} \right) = -\frac{p_1p_2}{4\pi\varepsilon_0} (-3x^{-4}) = \frac{3p_1p_2}{4\pi\varepsilon_0 x^4}$.However,for two parallel dipoles placed side-by-side (equatorial configuration),the force is attractive and its magnitude is $F = \frac{6p_1p_2}{4\pi\varepsilon_0 x^4}$.

Two point dipoles of dipole moments $\vec{p}_1$ and $\vec{p}_2$ are at a distance $x$ from each other,and $\vec{p}_1 \parallel \vec{p}_2$. The force between the dipoles is:

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