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Field at Q due to charge $+ 10 \mu C$ at B

$\qquad= \large\frac{10^{-5}\;C}{4 \pi (8.854 \times 10^{-12} C^{2} N^{-1} m^{-2})} \times \large\frac{1}{[15^2+(0.25)^2] \times 10^{-4} m^2}$

$\qquad= 3.99 × 10^6\; N C^{–1}$ along BQ

Field at Q due to charge $–10\; \mu \;C$ at $A$

$\qquad= \large\frac{10^{-5}\;C}{4 \pi (8.854 \times 10^{-12} C^{2} N^{-1} m^{-2})} \times \large\frac{1}{[15^2+(0.25)^2] \times 10^{-4} m^2}$

$\qquad= 3.99 × 10^6\; N C^{–1}$ along QA

Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is

$\qquad= 2 \times \large\frac{0.25 }{\sqrt {15^2 +(0.25)62 } } $$ \times 3.99 \times 10^6 \;NC^{-1} $ along BA

$\qquad= 1.33 \times 10^5 \;NC^{-1}$

The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again the result agrees with that obtained before.

Hence C is the correct answer.

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