Example

Question:

Two charges \( \pm 10\,\mu\text{C} \) are placed \( 5.0\,\text{mm} \) apart. Determine the electric field at (a) a point \( P \) on the axis of the dipole \( 15\,\text{cm} \) away from its centre \( O \) on the side of the positive charge, as shown in Fig. 1.18(a), and (b) a point \( Q \), \( 15\,\text{cm} \) away from \( O \) on a line passing through \( O \) and normal to the axis of the dipole, as shown in Fig. 1.18(b).

Solution:

(a) Field at \( P \) due to charge \( +10\,\mu\text{C} \):
\[ E_{BP} = \frac{10^{-5}\,\text{C}}{4\pi(8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}) \times (15+0.025)\times10^{-2}\,\text{m}^2} = 4.13 \times 10^6\,\text{N\,C}^{-1} \] along BP.
Field at \( P \) due to charge \( -10\,\mu\text{C} \):
\[ E_{AP} = \frac{10^{-5}\,\text{C}}{4\pi(8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}) \times (15+0.025)\times10^{-2}\,\text{m}^2} = 3.86 \times 10^6\,\text{N\,C}^{-1} \] along PA.
The resultant electric field at \( P \) is \[ E_P = 2.7 \times 10^5\,\text{N\,C}^{-1} \] For a dipole, at far-away point on the axis, the field magnitude is \[ E = \frac{2p}{4\pi\varepsilon_0 r^3} \] where \( p = 2aq \) is the magnitude of the dipole moment, \( a \) is half the separation and \( q \) is charge. \[ p = 10^{-5}\,\text{C} \times 5.0 \times 10^{-3}\,\text{m} = 5 \times 10^{-8}\,\text{C\,m} \] \[ E = \frac{2 \times 5 \times 10^{-8}\,\text{C\,m}}{4\pi(8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}) \times (15\times 10^{-2}\,\text{m})^3} = 2.6 \times 10^5\,\text{N\,C}^{-1} \] along the dipole moment direction AB.

(b) Field at \( Q \) due to charge \( +10\,\mu\text{C} \) at B: \[ E_{BQ} = \frac{10^{-5}\,\text{C}}{4\pi(8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}) \times [15^2+(0.025)^2] \times 10^{-4}\,\text{m}^2} = 3.99 \times 10^6\,\text{N\,C}^{-1} \] along BQ.
Field at \( Q \) due to charge \( -10\,\mu\text{C} \) at A: \[ E_{AQ} = \frac{10^{-5}\,\text{C}}{4\pi(8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}) \times [15^2+(0.025)^2] \times 10^{-4}\,\text{m}^2} = 3.99 \times 10^6\,\text{N\,C}^{-1} \] along QA.
The resultant along BA is \[ E_Q = 2 \times \frac{0.25}{15} \times 3.99 \times 10^6\,\text{N\,C}^{-1} = 1.33 \times 10^5\,\text{N\,C}^{-1} \] For a dipole, at a point on the normal to the axis, \[ E = \frac{p}{4\pi\varepsilon_0 r^3} \] \[ = \frac{5 \times 10^{-8}\,\text{C\,m}}{4\pi(8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}) \times (15 \times 10^{-2}\,\text{m})^3} = 1.33 \times 10^5\,\text{N\,C}^{-1} \] This result agrees with the earlier calculation.