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An electric dipole consists of two equal and opposite point charges +Q and –Q, separated by a distance 2l. P is a point collinear with the charges such that its distance from the positive charge is half of its distance from the negative charge. Magnitude of the electric field intensity at P is

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$\frac{1}{4 {πε}_{0}} \frac{3 Q}{16 l^{2}}$

$\frac{1}{4 {πε}_{0}} \frac{Q}{16 l^{2}}$

$\frac{1}{4 {πε}_{0}} \frac{3 Q}{4 l^{2}}$

$\frac{1}{4 {πε}_{0}} \frac{Q}{l^{2}}$

answer is A.

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Detailed Solution

Let point ‘P’ is at a distance ‘x’ from +ve charge the distance of P from –ve charge is 2l + x. Here $x = \frac{1}{2} (2 l + x) \Rightarrow x = 2 l$

$E = \frac{1}{4 {πε}_{0}} (\frac{Q}{x^{2}} - \frac{Q}{(2 l + x)^{2}}) = \frac{1}{4 {πε}_{0}} (\frac{Q}{4 l^{2}} - \frac{Q}{16 l^{2}}) = \frac{1}{4 {πε}_{0}} \frac{3 Q}{16 l^{2}}$

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