Physics · Ch 1 — Electrostatics
Electric field due to a dipole
Electric field due to a dipole
The electric dipole itself produces an electric field in the surrounding space, obtained by superposing the individual point-charge fields of its +q and -q charges. Two special cases are worked out in full. Case (i), the axial line: for a point C on the line through both charges, extended a distance r from the dipole's midpoint O (with r much greater than the charge separation 2a), the fields of +q and -q add (since C lies outside the pair, both contributions ultimately point the same way along the axis) to give E_axial = (1/4 pi epsilon0) 2pr/(r^2-a^2)^2, which for r much greater than a reduces to E_axial is approximately 2kp/r^3, directed parallel to the dipole moment p. Case (ii), the equatorial line (the plane perpendicular to the dipole axis through its midpoint): for a point P on this perpendicular bisector, the two point-charge fields have equal magnitude by symmetry, and their components perpendicular to the axis cancel while their components along the axis add, giving E_equatorial = (1/4 pi epsilon0) p/(r^2+a^2)^(3/2), which for r much greater than a reduces to E_equatorial is approximately kp/r^3, directed anti-parallel to p. Comparing the two results at the same large distance r, the axial field is exactly twice as strong as the equatorial field (2kp/r^3 versus kp/r^3), and both fall off as the inverse cube of distance rather than the invers …
What this figure shows. A dipole -q at one end and +q at the other, separated by 2a, is drawn on the x-axis with midpoint O at the origin, and a field point C is marked further out along the same axis at distance r from O. Two field vectors are drawn at C -- one due to +q pointing away from it (in the +x direction, since C is on the side of the positive charge) and a smaller one due to -q pointing toward it (also, after the geometry, ending up along +x) -- setting up the vector addition that gives the net ax …
What this figure shows. The two individual point-charge field vectors from Figure 1.15 are redrawn along a single axis and added algebraically (since both ultimately point in the same direction along the axis for a point outside the charges), giving the single net resultant field vector E_axial at point C. It is the completion, step by step, of the superposition calculation whose final closed-form result is the standard axial dipole-field formula, E = (1/4 pi epsilon0)(2pr/(r^2-a^2)^2), which simplifies to 2kp/r^3 when r is much larger …
Worked out. For a field point C on the axial line at distance r from the dipole's midpoint O (with r taken much larger than the charge separation 2a), the field due to +q alone has magnitude k q/(r-a)^2 pointing away from the dipole along the axis, and the field due to -q alone has magnitude k q/(r+a)^2 pointing toward -q, i.e. also along the same axis direction at C. Adding these two collinear vectors and simplifying using the difference of squares (r-a)(r+a) = r^2-a^2 gives the exact result E_axial = (1/4 pi epsilon0) x 2pr/(r^2-a^2)^2, directed along the dipole moment p. For a short dipole viewed from far away (r much greater than a), the a^2 term in the denominator is negligible compared with r^2, and the formula reduces to the widely used approximation E_axial is approximately equal to 2kp/r^3, showing that a dipole's axial field falls off as the inverse cube of distance, faster than a single …
What this figure shows. A dipole -q, +q sits on the x-axis with midpoint O, and a field point P is marked on the y-axis, equidistant from both charges (the equatorial line, the perpendicular bisector of the dipole). Two equal-length field vectors are drawn from P, one pointing away from +q and one pointing toward -q; their vertical (y-direction) components cancel by symmetry while their horizontal (x-direction, anti-parallel to p) components add, leaving a net field directed exactly opposite to the dipole …
Worked out. For a field point P on the equatorial line, equidistant from both charges at distance sqrt(r^2+a^2) from each, the two individual point-charge fields have equal magnitude k q/(r^2+a^2), but by the symmetry of the geometry their components perpendicular to the dipole axis exactly cancel while their components parallel to the axis (both pointing from +q's side toward -q's side, i.e. anti-parallel to p) add together. Working through the geometry gives the exact result E_equatorial = (1/4 pi epsilon0) x p/(r^2+a^2)^(3/2), directed opposite to p, which for r much larger than a simplifies to E_equatorial is approximately equal to kp/r^3 -- exactly half the magnitude of the axial field at the same distance r, and pointing in the opposite sense (anti-parall …