Charges –q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the midpoint of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >>2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P’ such that the OP’ $= (\frac{y}{3}),$ force on Q will be close to $(\frac{y}{3} > > 2 a)$ is $N \times F$ find the value of $N$

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answer is 27.

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Detailed Solution

Electric field due to dipole on equatorial plane, $\vec{E} = - k \frac{\vec{p}}{r^{3}}$

At point P, ${\vec{F}}_{p} = - k \frac{\vec{p}}{y^{3}} Q ............... (i)$

At point $P^{'}, {\vec{F}}_{p} = - (- k \frac{\vec{p}}{{(y / 3)}^{3}} Q) ............... (i i)$

From equation (i) and (ii), ${\vec{F}}_{p^{'}} = 27 {\vec{F}}_{p^{'}} = 27 \vec{F}$

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