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(D) The electric field $E$ due to an electric dipole at an axial position (end-on position) at a distance $r$ is given by $E = \frac{1}{4\pi\epsilon_0

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$F/8$

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(D) The electric field $E$ due to an electric dipole at an axial position (end-on position) at a distance $r$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{2pr}{ (r^2 - a^2)^2 }$. For a point at a large distance $(r \gg a)$,this simplifies to $E = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$.Since the force $F$ on a charge $q$ is $F = qE$,we have $F = q \left( \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} \right)$.This implies that $F \propto \frac{1}{r^3}$.If the distance $r$ is doubled $(r' = 2r)$,the new force $F'$ will be $F' \propto \frac{1}{(2r)^3} = \frac{1}{8r^3}$.Therefore,$F' = \frac{F}{8}$.

$A$ given charge is situated at a certain distance from an electric dipole in the end-on position experiences a force $F$. If the distance of the charge is doubled,the force acting on the charge will be

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