Physics · Ch 10 — Electrostatics

Electric Intensity at a Point due to an Electric Dipole

10.9.2

Electric Intensity at a Point due to an Electric Dipole

This section finds the electric field intensity due to a dipole (charges $-q$ at A and $+q$ at B, separated by $2l$ , with dipole moment $p=q(2l)$ ) at two specific, important points in space -- one on its axial line, and one on its equatorial line.

Case 1: At a point on the axial line. Let P be a point on the dipole's extended axis, at distance r from the dipole's centre C (Fig. 10.20), so that $AP=(r+l)$ and $BP=(r-l)$ . The field at P due to the $-q$ charge at A has magnitude $E_A=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{(r+l)^2}$ , directed back toward A; the field at P due to the $+q$ charge at B has magnitude $E_B=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{(r-l)^2}$ , directed away from B. Since P is nearer to B than to A, $E_B>E_A$ , and both fields point in the SAME overall direction along the axis, so the resultant field $\vec{E}_a=\vec{E}_B+\vec{E}_A$ (algebraically, $E_B-E_A$ for the opposing-direction convention) simplifies, after combining the two fractions over a common denominator, to $E_a=\frac{1}{4\pi\epsilon_0}\frac{2pr}{(r^2-l^2)^2}.$ For points far from the dipole compared to its own size, $r\gg l$ , the $l^2$ term becomes negligible next to $r^2$ , giving the simpler, widely-used approximate result $E_a\approx\frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}\qquad(r\gg l),$ directed ALONG the dipole moment $\vec{p}$ , i.e. from the negative charge toward the positive charge. …

Figure 10.20Fig. 10.20: Electric field of a dipole at a point on its axial line

What this figure shows. A dipole with charge $-q$ at point A and $+q$ at point B, centred at C, separated by $2l$ . A point P is marked on the extended axial line beyond B, at distance r from the centre C (so $AP=r+l$ and $BP=r-l$ ). Two field vectors are drawn at P: $\vec{E}_A$ , due to the $-q$ charge at A, pointing back TOWARD A (since the source is negative); and $\vec{E}_B$ , due to the $+q$ charge at B, pointing AWAY from B (along the axis, in the same general direction as $\vec{E}_A$ here since P is beyond B). The figure sets up the vector addition $\vec{E}_a=\vec{E}_A+\vec{E}_B$ used to derive the axial-field formula, showing the resultant field points along the axis in the direction of the d …

Figure 10.21Fig. 10.21 (a)-(c): Electric field of a dipole at a point on its equatorial line, its components, and its direction

What this figure shows. Panel (a) shows a dipole with $-q$ at A and $+q$ at B, and a point P on the equatorial line (the perpendicular bisector of AB), at distance r from the centre; field vectors $\vec{E}_A$ (pointing from P toward A, since A is negative) and $\vec{E}_B$ (pointing from B toward P and beyond, since B is positive) are drawn, both of equal magnitude since $AP=BP$ . Panel (b) resolves these two vectors into components along the equatorial line (y-axis) and along a direction parallel to the dipole axis (x-axis): the y-components (perpendicular-to-axis components) of $\vec{E}_A$ and $\vec{E}_B$ are equal and opposite and CANCEL, while their x-components (along-the-axis components) are equal and point in the SAME direction, adding to give the net equatorial field. Panel (c) shows the final resultant field vector at P directed ANTI-PARALLEL to the dipole moment $\vec{p}$ (i.e. pointing from the positive-charge end toward the negative-charge end, the opposite …

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