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Field at P due to charge $+10 \mu C$

$\qquad= \large\frac{19^{-5}\;C}{4 \pi (8.854 \times 10^{-12} C^{2} N^{-1} m^{-2})} \times \large\frac{1}{(15-.25)^2 \times 10^{-4} m^2}$

$\qquad=4.13 \times 10^5 NC^{-1} $ along BP.

$\qquad= \large\frac{19^{-5}\;C}{4 \pi (8.854 \times 10^{-12} C^{2} N^{-1} m^{-2})} \times \large\frac{1}{(15+.25)^2 \times 10^{-4} m^2}$

$\qquad= 3.86 \times 10^5 NC^{-1} $ along PA.

The resultant electric field at P due to the two charges at A and B is = $2.7 \times 10^ 5 N C^{–1}$ along BP.

In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole.

For a dipole consisting of charges $\pm q, 2a$ distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude

$E= \large\frac{2p}{4 \pi \in_0 r^3}$

where $p = 2a$ q is the magnitude of the dipole moment.

The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from –q to q). Here,

$p =10^{–5}\; C \times 5 \times 10^{–3} m = 5 \times 10^{–8} C m$

Therefore ,

$\qquad= \large\frac{2 \times 5 \times 10^{-8}\;Cm}{4 \pi (8.854 \times 10^{-12} C^{2} N^{-1} m^{-2})} \times \large\frac{1}{(15)^3+10^{-6} m^3}$$=2.6 \times 10^5 NC^{-1}$

Along the dipole moment direction AB, which is close to the result obtained earlier.

Hence A is the correct answer.

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