Question 238 of 311

Two shorts dipoles $(A, B), A$ having charges $\pm 2 \mu \mathrm{C}$ and length 1 cm and $B$ having charges $\pm 4 \mu \mathrm{C}$ and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point $P$, equi-distant from the centres of both dipoles is $\_\_\_\_$ N/C.

A$\frac{9}{16} \sqrt{2} \times 10^{4}$B$9 \sqrt{2} \times 10^{4}$C$4.5 \sqrt{2} \times 10^{4}$D$\frac{9}{16} \sqrt{2} \times 10^{5}$

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Answer: (A) $\frac{9}{16} \sqrt{2} \times 10^{4}$

$\vec{E}_2 = -\dfrac{KP_2}{r^3}$ ; $\vec{E}_1 = -\dfrac{2KP_1}{r^3}$
$P_1 = 2 \times 10^{-6} \times 10^{-2} = 2 \times 10^{-8}$
$P_2 = 4 \times 10^{-6} \times 10^{-2} = 4 \times 10^{-8}$
$\vec{E}_{net} = \dfrac{2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{(0.4)^3} \hat{i} - \dfrac{9 \times 10^9 \times 4 \times 10^{-8}}{(0.4)^3} \hat{j}$
$\vec{E}_{net} = \dfrac{9 \times 10^9 \times 4 \times 10^{-8}}{(0.4)^3} [\hat{i} - \hat{j}]$
$|\vec{E}_{net}| = \dfrac{9 \times 10^4}{16} (\sqrt{2})$