Electrostatics Part 3

Concept:Electric field due to a dipole depends on orientation: axial field is $E_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$ and equatorial field is $E_{\text{eq}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$.Explanation:Dipole moment: $p = q \times 2l$.For dipole A: $q_A = 2\,\mu\mathrm{C} = 2\times10^{-6}\,\mathrm{C}$, $2l = 1\,\mathrm{cm} = 0.01\,\mathrm{m}$.$p_A = (2\times10^{-6}) \times 0.01 = 2\times10^{-8}\,\mathrm{C\!\cdot\!m}$.For dipole B: $q_B = 4\,\mu\mathrm{C} = 4\times10^{-6}\,\mathrm{C}$, $2l = 0.01\,\mathrm{m}$.$p_B = (4\times10^{-6}) \times 0.01 = 4\times10^{-8}\,\mathrm{C\!\cdot\!m}$.Centres are $80\,\mathrm{cm}$ apart. Point P is equidistant, so $r = 40\,\mathrm{cm} = 0.4\,\mathrm{m}$.P lies on the axial line of dipole A and the equatorial line of dipole B.Field due to A (axial): $E_A = \frac{1}{4\pi\epsilon_0} \frac{2p_A}{r^3} = \frac{9\times10^9 \times 2 \times 2\times10^{-8}}{(0.4)^3} = \frac{360}{0.064} = 5625\,\mathrm{N/C}$, horizontal.Field due to B (equatorial): $E_B = \frac{1}{4\pi\epsilon_0} \frac{p_B}{r^3} = \frac{9\times10^9 \times 4\times10^{-8}}{0.064} = \frac{360}{0.064} = 5625\,\mathrm{N/C}$, vertical.Since $E_A$ and $E_B$ are perpendicular and equal in magnitude: $E_{\text{net}} = \sqrt{E_A^2 + E_B^2} = E\sqrt{2} = 5625\sqrt{2}\,\mathrm{N/C}$.$5625 = \frac{9}{16} \times 10^4$, so $E_{\text{net}} = \frac{9}{16}\sqrt{2}\times10^4\,\mathrm{N/C}$.Answer:Option B: $\frac{9}{16}\sqrt{2}\times10^4$ N/C.