Field at distance r from centre on the equatorial plane. Direction is antiparallel to p̂. Magnitude is half of axial field at same r (far field).
Setup
Same dipole as before: − q -q − q ( 0 , − a , 0 ) (0, -a, 0) ( 0 , − a , 0 )
Let me use the standard orientation: dipole along the x-axis, + q +q + q ( a , 0 ) (a, 0) ( a , 0 ) − q -q − q ( − a , 0 ) (-a, 0) ( − a , 0 )
Consider point P at ( 0 , r ) (0, r) ( 0 , r )
d = r 2 + a 2 d = \sqrt{r^2 + a^2} d = r 2 + a 2 Fields from each charge
By symmetry, the x-components of E + E_+ E + E − E_- E −
The y-components also cancel... wait, let me redo.
E ⃗ + \vec{E}_+ E + + q +q + q ( a , 0 ) (a,0) ( a , 0 ) ( 0 , r ) (0,r) ( 0 , r ) ( − a , r ) / d (-a, r)/d ( − a , r ) / d
E ⃗ + = q 4 π ε 0 d 2 ⋅ ( − a , r ) d \vec{E}_+ = \frac{q}{4\pi\varepsilon_0 d^2} \cdot \frac{(-a, r)}{d} E + = 4 π ε 0 d 2 q ⋅ d ( − a , r ) E ⃗ − \vec{E}_- E − − q -q − q ( − a , 0 ) (-a,0) ( − a , 0 ) ( 0 , r ) (0,r) ( 0 , r ) toward − q -q − q ( − a , − r ) ⋅ ( − 1 ) (−a, −r)\cdot(-1) ( − a , − r ) ⋅ ( − 1 )
Let me be careful. Field due to − q -q − q toward − q -q − q
Direction from P to − q -q − q ( − a , 0 ) − ( 0 , r ) = ( − a , − r ) (-a, 0) - (0, r) = (-a, -r) ( − a , 0 ) − ( 0 , r ) = ( − a , − r ) ( − a , − r ) / d (-a,-r)/d ( − a , − r ) / d
So:
E ⃗ − = q 4 π ε 0 d 2 ⋅ ( − a , − r ) d \vec{E}_- = \frac{q}{4\pi\varepsilon_0 d^2} \cdot \frac{(-a, -r)}{d} E − = 4 π ε 0 d 2 q ⋅ d ( − a , − r ) Net field
E ⃗ = E ⃗ + + E ⃗ − = q 4 π ε 0 d 3 [ ( − a , r ) + ( − a , − r ) ] = q 4 π ε 0 d 3 ( − 2 a , 0 ) \vec{E} = \vec{E}_+ + \vec{E}_- = \frac{q}{4\pi\varepsilon_0 d^3}\left[(-a,r) + (-a,-r)\right] = \frac{q}{4\pi\varepsilon_0 d^3}(-2a, 0) E = E + + E − = 4 π ε 0 d 3 q [ ( − a , r ) + ( − a , − r ) ] = 4 π ε 0 d 3 q ( − 2 a , 0 ) The y-components cancel. The x-component:
E = − 2 q a 4 π ε 0 ( r 2 + a 2 ) 3 / 2 = − p 4 π ε 0 ( r 2 + a 2 ) 3 / 2 E = \frac{-2qa}{4\pi\varepsilon_0 (r^2+a^2)^{3/2}} = \frac{-p}{4\pi\varepsilon_0(r^2+a^2)^{3/2}} E = 4 π ε 0 ( r 2 + a 2 ) 3/2 − 2 q a = 4 π ε 0 ( r 2 + a 2 ) 3/2 − p The negative sign means the field is in the − x -x − x antiparallel to p ⃗ \vec{p} p .
E ⃗ e q = − p 4 π ε 0 ( r 2 + a 2 ) 3 / 2 p ^ \boxed{\vec{E}_{eq} = -\frac{p}{4\pi\varepsilon_0(r^2+a^2)^{3/2}}\hat{p}} E e q = − 4 π ε 0 ( r 2 + a 2 ) 3/2 p p ^ Far-field approximation (r ≫ a r \gg a r ≫ a
E ≈ p 4 π ε 0 r 3 E \approx \frac{p}{4\pi\varepsilon_0 r^3} E ≈ 4 π ε 0 r 3 p
Remember
Compare with the axial field: $E_{axial} \approx 2p/(4\pi\varepsilon_0 r^3)$. At the same distance, the axial field is twice the equatorial field. Both directions are antiparallel to each other at a given $r$.