Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 3.1.4

Electric dipole

Show that the electric field intensity at point $P (x, y, z)$ is

E (P) = \frac{q}{4 π ε _{0} R ^{3}} [3 \frac{R \cdot d}{R ^{2}} R - d]

for $d ≪ R$ .

Try it yourself first.

Strategy

Superposition of two point-charge fields (3.1.25) gives the exact field at once. The work is in the approximation $d ≪ R$ : expand the two denominators to first order in $d / R$ with $(1 + x)^{n} ≃ 1 + n x$ and collect terms.

Let $d$ be the vector from $- q$ to $+ q$ , with the origin at the midpoint, so that $+ q$ is at $d /2$ and $- q$ at $- d /2$ , and let $R$ be the position vector of P. The displacement vectors from $+ q$ and from $- q$ to P are

R_{1} = R - \frac{d}{2} and R_{2} = R + \frac{d}{2},

respectively, and the superposition principle (3.1.25) gives

E (P) = \frac{q}{4 π ε _{0}} ⎩ ⎨ ⎧ \frac{R - \frac{d}{2}}{R - \frac{d}{2} ^{3}} - \frac{R + \frac{d}{2}}{R + \frac{d}{2} ^{3}} ⎭ ⎬ ⎫ .

This is exact. To simplify it for $d ≪ R$ we expand the denominators, neglecting $d^{2} /4$ against $R^{2}$ and using $(1 + x)^{n} ≃ 1 + n x$ for $∣ x ∣ ≪ 1$ :

R - \frac{d}{2}^{- 3} = [(R - \frac{d}{2}) \cdot (R - \frac{d}{2})]^{- 3/2} = [R^{2} - R \cdot d + \frac{d ^{2}}{4}]^{- 3/2} ≃ R^{- 3} [1 - \frac{R \cdot d}{R ^{2}}]^{- 3/2} ≃ R^{- 3} [1 + \frac{3}{2} \frac{R \cdot d}{R ^{2}}],

and in the same way

R + \frac{d}{2}^{- 3} ≃ R^{- 3} [1 - \frac{3}{2} \frac{R \cdot d}{R ^{2}}] .

Substituting these,

E ≃ \frac{q}{4 π ε _{0} R ^{3}} {[1 + \frac{3}{2} \frac{R \cdot d}{R ^{2}}] (R - \frac{d}{2}) - [1 - \frac{3}{2} \frac{R \cdot d}{R ^{2}}] (R + \frac{d}{2})} = \frac{q}{4 π ε _{0} R ^{3}} {R - \frac{d}{2} + \frac{3}{2} \frac{R \cdot d}{R ^{2}} R - \frac{3}{4} \frac{R \cdot d}{R ^{2}} d - R - \frac{d}{2} + \frac{3}{2} \frac{R \cdot d}{R ^{2}} R + \frac{3}{4} \frac{R \cdot d}{R ^{2}} d} .

The terms $\pm R$ and $\pm \frac{3}{4} (R \cdot d / R^{2}) d$ cancel, the two $- d /2$ add to $- d$ , and the two $\frac{3}{2} (R \cdot d / R^{2}) R$ add to $3 (R \cdot d / R^{2}) R$ . Hence, for $d ≪ R$ ,

Answer

E (P) ≃ \frac{q}{4 π ε _{0} R ^{3}} [3 \frac{R \cdot d}{R ^{2}} R - d] [V/m]

Interpretation

With $d = d a_{z}$ and spherical coordinates centred on the dipole ( $R = r$ ), we have $R \cdot d = r d cos θ$ and $a_{z} = cos θ a_{r} - sin θ a_{θ}$ , and the result becomes the familiar spherical form

E (P) ≃ \frac{q d}{4 π ε _{0} r ^{3}} (2 cos θ a_{r} + sin θ a_{θ}) .

Only the product $p = q d$ , the dipole moment, appears. The dipole field falls off as $1/ r^{3}$ , one power faster than the field of a single charge, because the net charge is zero and the fields of $+ q$ and $- q$ almost cancel far away (compare Figure 3.1.21a). Dipoles return when we study polarisation in Section 3.4.

Check

On the axis of the dipole ( $R ∥ d$ , $R = r$ ) the approximate result gives $E = 2 q d / (4 π ε_{0} r^{3})$ , while the exact superposition gives

E = \frac{q}{4 π ε _{0}} [\frac{1}{( r - d /2 ) ^{2}} - \frac{1}{( r + d /2 ) ^{2}}] = \frac{q}{4 π ε _{0}} \frac{2 r d}{( r ^{2} - d ^{2} /4 ) ^{2}} = \frac{2 q d}{4 π ε _{0} r ^{3}} \frac{1}{( 1 - d ^{2} /4 r ^{2} ) ^{2}} .

The last factor tends to $1$ for $d ≪ r$ . It is $1.14$ at $r = 2 d$ , $1.02$ at $r = 5 d$ and $1.005$ at $r = 10 d$ , so the approximation is already good a few dipole lengths away. Units: $q d / (ε_{0} r^{3})$ is $C m / (F/m \cdot m^{3}) = V/m$ .

Home exercise 3.1.4: Electric dipole