Deriving the Far Field of a Dipole and the Field of a Charged Arc
I Key FormulasKey Insight: The expansion is in powers of A / r A/r A / r 1 / r 1/r 1/ r E x E_x E x r 5 r^5 r 5 1 / r 3 1/r^3 1/ r 3 x y / r 2 xy/r^2 x y / r 2 − 1 2 -\tfrac{1}{2} − 2 1 1 2 \tfrac{1}{2} 2 1
II Visualization 1 — Exact Field Against the Dipole FormulaFar from the charges, the two-charge field and p 4 π ε 0 r 3 ( 2 cos θ r ^ + sin θ θ ^ ) \frac{p}{4\pi\varepsilon_0 r^3}(2\cos\theta\,\hat{\mathbf{r}} + \sin\theta\,\hat{\boldsymbol{\theta}}) 4 π ε 0 r 3 p ( 2 cos θ r ^ + sin θ θ ^ ) ( A / r ) 2 (A/r)^2 ( A / r ) 2
III Visualization 2 — Opening the ArcAt the centre only E x E_x E x 2 sin θ 0 2\sin\theta_0 2 sin θ 0 λ 4 π ε 0 R \frac{\lambda}{4\pi\varepsilon_0 R} 4 π ε 0 R λ
IV Quiz Questions
Problem 1 · On the Dipole Axis
Given: A dipole of moment p = 6.0 × 10 − 30 C m p = 6.0\times10^{-30}\ \text{C}\,\text{m} p = 6.0 × 1 0 − 30 C m 1 4 π ε 0 = 9.0 × 10 9 N m 2 / C 2 \frac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2 4 π ε 0 1 = 9.0 × 1 0 9 N m 2 / C 2 Find the field magnitude on its axis (θ = 0 \theta = 0 θ = 0 r = 2.0 × 10 − 9 m r = 2.0\times10^{-9}\ \text{m} r = 2.0 × 1 0 − 9 m
a 6.8 × 10 6 N/C 6.8\times10^{6}\ \text{N/C} 6.8 × 1 0 6 N/C b 2.7 × 10 − 2 N/C 2.7\times10^{-2}\ \text{N/C} 2.7 × 1 0 − 2 N/C c 2.7 × 10 7 N/C 2.7\times10^{7}\ \text{N/C} 2.7 × 1 0 7 N/C d 1.35 × 10 7 N/C 1.35\times10^{7}\ \text{N/C} 1.35 × 1 0 7 N/C
✅ Correct! At
θ = 0 \theta = 0 θ = 0 ,
E θ = 0 E_\theta = 0 E θ = 0 and
E r = 2 p / ( 4 π ε 0 r 3 ) E_r = 2p/(4\pi\varepsilon_0 r^3) E r = 2 p / ( 4 π ε 0 r 3 ) .
❌ Not quite. p / ( 4 π ε 0 r 3 ) p/(4\pi\varepsilon_0 r^3) p / ( 4 π ε 0 r 3 ) is the bisector value; on the axis
2 cos θ = 2 2\cos\theta = 2 2 cos θ = 2 .
❌ Not quite. The dipole field falls as
1 / r 3 1/r^3 1/ r 3 , not
1 / r 2 1/r^2 1/ r 2 .
❌ Not quite. 3 cos 2 θ + 1 = 2 \sqrt{3\cos^2\theta + 1} = 2 3 cos 2 θ + 1 = 2 at
θ = 0 \theta = 0 θ = 0 , not
4 4 4 .
❌ Not quite. Use
E r = 2 p cos θ / ( 4 π ε 0 r 3 ) E_r = 2p\cos\theta/(4\pi\varepsilon_0 r^3) E r = 2 p cos θ / ( 4 π ε 0 r 3 ) .
Show solution
E = 2 p 4 π ε 0 r 3 = 2 ( 9.0 × 10 9 ) ( 6.0 × 10 − 30 ) ( 2.0 × 10 − 9 ) 3 = 1.08 × 10 − 19 8.0 × 10 − 27 = 1.35 × 10 7 N/C . E = \frac{2p}{4\pi\varepsilon_0 r^3} = \frac{2(9.0\times10^{9})(6.0\times10^{-30})}{(2.0\times10^{-9})^3} = \frac{1.08\times10^{-19}}{8.0\times10^{-27}} = 1.35\times10^{7}\ \text{N/C}. E = 4 π ε 0 r 3 2 p = ( 2.0 × 1 0 − 9 ) 3 2 ( 9.0 × 1 0 9 ) ( 6.0 × 1 0 − 30 ) = 8.0 × 1 0 − 27 1.08 × 1 0 − 19 = 1.35 × 1 0 7 N/C .
Problem 2 · On the Bisector
Given: the dipole points along + j ^ +\hat{\mathbf{j}} + j ^ θ = 90 ° \theta = 90\degree θ = 90° Find the direction of the field at P.
a Along − j ^ -\hat{\mathbf{j}} − j ^ p ⃗ \vec{\mathbf{p}} p b Along + j ^ +\hat{\mathbf{j}} + j ^ p ⃗ \vec{\mathbf{p}} p c Along r ^ \hat{\mathbf{r}} r ^ d There is no field: the charges cancel
✅ Correct! E r = 0 E_r = 0 E r = 0 and
E θ = p / ( 4 π ε 0 r 3 ) E_\theta = p/(4\pi\varepsilon_0 r^3) E θ = p / ( 4 π ε 0 r 3 ) ; at
θ = 90 ° \theta = 90\degree θ = 90° ,
θ ^ = − j ^ \hat{\boldsymbol{\theta}} = -\hat{\mathbf{j}} θ ^ = − j ^ .
❌ Not quite. That is the direction on the axis. On the bisector
θ ^ = cos 90 ° i ^ − sin 90 ° j ^ = − j ^ \hat{\boldsymbol{\theta}} = \cos 90\degree\,\hat{\mathbf{i}} - \sin 90\degree\,\hat{\mathbf{j}} = -\hat{\mathbf{j}} θ ^ = cos 90° i ^ − sin 90° j ^ = − j ^ .
❌ Not quite. E r = 2 p cos 90 ° / ( 4 π ε 0 r 3 ) = 0 E_r = 2p\cos 90\degree/(4\pi\varepsilon_0 r^3) = 0 E r = 2 p cos 90°/ ( 4 π ε 0 r 3 ) = 0 there.
❌ Not quite. Only the zeroth-order terms cancel; the first-order terms in
A / r A/r A / r leave
p / ( 4 π ε 0 r 3 ) p/(4\pi\varepsilon_0 r^3) p / ( 4 π ε 0 r 3 ) .
❌ Not quite. Evaluate
E r E_r E r and
E θ E_\theta E θ at
θ = 90 ° \theta = 90\degree θ = 90° .
Show solution
At θ = 90 ° \theta = 90\degree θ = 90° E r = 0 E_r = 0 E r = 0 E θ = p 4 π ε 0 r 3 E_\theta = \dfrac{p}{4\pi\varepsilon_0 r^3} E θ = 4 π ε 0 r 3 p θ \theta θ y y y
θ ^ = cos 90 ° i ^ − sin 90 ° j ^ = − j ^ , \hat{\boldsymbol{\theta}} = \cos 90\degree\,\hat{\mathbf{i}} - \sin 90\degree\,\hat{\mathbf{j}} = -\hat{\mathbf{j}}, θ ^ = cos 90° i ^ − sin 90° j ^ = − j ^ , so the field points opposite to p ⃗ \vec{\mathbf{p}} p r r r
Problem 3 · Off the Axis
Given: the dipole of Problem 1, p = 6.0 × 10 − 30 C m p = 6.0\times10^{-30}\ \text{C}\,\text{m} p = 6.0 × 1 0 − 30 C m r = 2.0 × 10 − 9 m r = 2.0\times10^{-9}\ \text{m} r = 2.0 × 1 0 − 9 m θ = 60 ° \theta = 60\degree θ = 60° Find ∣ E ⃗ ∣ |\vec{\mathbf{E}}| ∣ E ∣
a 1.3 × 10 7 N/C 1.3\times10^{7}\ \text{N/C} 1.3 × 1 0 7 N/C b 1.2 × 10 7 N/C 1.2\times10^{7}\ \text{N/C} 1.2 × 1 0 7 N/C c 8.9 × 10 6 N/C 8.9\times10^{6}\ \text{N/C} 8.9 × 1 0 6 N/C d 6.8 × 10 6 N/C 6.8\times10^{6}\ \text{N/C} 6.8 × 1 0 6 N/C
✅ Correct! 3 cos 2 60 ° + 1 = 1.75 = 1.32 \sqrt{3\cos^2 60\degree + 1} = \sqrt{1.75} = 1.32 3 cos 2 60° + 1 = 1.75 = 1.32 times
6.75 × 10 6 N/C 6.75\times10^{6}\ \text{N/C} 6.75 × 1 0 6 N/C .
❌ Not quite. E r E_r E r and
E θ E_\theta E θ are perpendicular; combine them with Pythagoras, not by adding.
❌ Not quite. That uses
θ = 30 ° \theta = 30\degree θ = 30° ; here
θ \theta θ is measured from the dipole axis and is
60 ° 60\degree 60° .
❌ Not quite. p / ( 4 π ε 0 r 3 ) p/(4\pi\varepsilon_0 r^3) p / ( 4 π ε 0 r 3 ) is only the scale; multiply by
3 cos 2 θ + 1 \sqrt{3\cos^2\theta + 1} 3 cos 2 θ + 1 .
❌ Not quite. Find
E r E_r E r and
E θ E_\theta E θ , then the magnitude.
Show solution
Scale: p 4 π ε 0 r 3 = ( 9.0 × 10 9 ) ( 6.0 × 10 − 30 ) 8.0 × 10 − 27 = 6.75 × 10 6 N/C \dfrac{p}{4\pi\varepsilon_0 r^3} = \dfrac{(9.0\times10^{9})(6.0\times10^{-30})}{8.0\times10^{-27}} = 6.75\times10^{6}\ \text{N/C} 4 π ε 0 r 3 p = 8.0 × 1 0 − 27 ( 9.0 × 1 0 9 ) ( 6.0 × 1 0 − 30 ) = 6.75 × 1 0 6 N/C
E r = 2 cos 60 ° × 6.75 × 10 6 = 6.75 × 10 6 , E θ = sin 60 ° × 6.75 × 10 6 = 5.85 × 10 6 N/C . E_r = 2\cos 60\degree \times 6.75\times10^{6} = 6.75\times10^{6},\qquad E_\theta = \sin 60\degree \times 6.75\times10^{6} = 5.85\times10^{6}\ \text{N/C}. E r = 2 cos 60° × 6.75 × 1 0 6 = 6.75 × 1 0 6 , E θ = sin 60° × 6.75 × 1 0 6 = 5.85 × 1 0 6 N/C . ∣ E ⃗ ∣ = E r 2 + E θ 2 = 6.75 × 10 6 1.75 = 8.9 × 10 6 N/C . |\vec{\mathbf{E}}| = \sqrt{E_r^2 + E_\theta^2} = 6.75\times10^{6}\sqrt{1.75} = 8.9\times10^{6}\ \text{N/C}. ∣ E ∣ = E r 2 + E θ 2 = 6.75 × 1 0 6 1.75 = 8.9 × 1 0 6 N/C .
Problem 4 · A Quarter-Circle Arc
Given: A quarter-circle arc (total angle 90 ° 90\degree 90° R = 0.10 m R = 0.10\ \text{m} R = 0.10 m λ = 2.0 nC/m \lambda = 2.0\ \text{nC/m} λ = 2.0 nC/m 1 4 π ε 0 = 9.0 × 10 9 N m 2 / C 2 \frac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2 4 π ε 0 1 = 9.0 × 1 0 9 N m 2 / C 2 Find the field magnitude at the centre.
a 360 N/C 360\ \text{N/C} 360 N/C b 255 N/C 255\ \text{N/C} 255 N/C c 283 N/C 283\ \text{N/C} 283 N/C d 127 N/C 127\ \text{N/C} 127 N/C
✅ Correct! θ 0 = 45 ° \theta_0 = 45\degree θ 0 = 45° is half the arc's angle, and
2 sin 45 ° = 1.414 2\sin 45\degree = 1.414 2 sin 45° = 1.414 .
❌ Not quite. θ 0 \theta_0 θ 0 is the half-angle: the arc runs from
− θ 0 -\theta_0 − θ 0 to
θ 0 \theta_0 θ 0 , so
θ 0 = 45 ° \theta_0 = 45\degree θ 0 = 45° , not
90 ° 90\degree 90° .
❌ Not quite. 2 θ 0 2\theta_0 2 θ 0 in place of
2 sin θ 0 2\sin\theta_0 2 sin θ 0 treats the arc as a point charge
Q Q Q ; a quarter circle is too long for that.
❌ Not quite. The two halves of the arc both contribute
sin θ 0 \sin\theta_0 sin θ 0 , giving the factor
2 2 2 .
❌ Not quite. Use
E = 2 λ sin θ 0 4 π ε 0 R E = \frac{2\lambda\sin\theta_0}{4\pi\varepsilon_0 R} E = 4 π ε 0 R 2 λ s i n θ 0 with the half-angle.
Show solution
E = 1 4 π ε 0 2 λ sin θ 0 R = ( 9.0 × 10 9 ) ( 2 ) ( 2.0 × 10 − 9 ) sin 45 ° 0.10 = 36 × 0.7071 0.10 = 255 N/C , E = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta_0}{R} = \frac{(9.0\times10^{9})(2)(2.0\times10^{-9})\sin 45\degree}{0.10} = \frac{36 \times 0.7071}{0.10} = 255\ \text{N/C}, E = 4 π ε 0 1 R 2 λ sin θ 0 = 0.10 ( 9.0 × 1 0 9 ) ( 2 ) ( 2.0 × 1 0 − 9 ) sin 45° = 0.10 36 × 0.7071 = 255 N/C , pointing from the arc's midpoint through the centre and beyond, away from the arc for λ > 0 \lambda > 0 λ > 0
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