A point particle of charge Q is located at P | vedclass

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Question

(C) The electric field at point $P$ due to dipole $1$ (axial position) is $E_1 = \frac{2kpr}{(r^2-a^2)^2}$,where $p = q(2a)$. However,looking at the f

Vedclass · Accepted Answer

$\frac{a}{r} \approx 0.5$

Vedclass · Answer

(C) The electric field at point $P$ due to dipole $1$ (axial position) is $E_1 = \frac{2kpr}{(r^2-a^2)^2}$,where $p = q(2a)$. However,looking at the forces from individual charges: The force from $+q$ of dipole $1$ is $F_1 = \frac{kqQ}{(r-a)^2}$ (repulsive) and from $-q$ is $F_2 = \frac{kqQ}{(r+a)^2}$ (attractive). The net force from dipole $1$ is $F_{net,1} = \frac{kqQ}{(r-a)^2} - \frac{kqQ}{(r+a)^2}$.For dipole $2$ (equatorial position),the electric field is $E_2 = \frac{kp}{(r^2+a^2)^{3/2}}$. The force from dipole $2$ on $Q$ is $F_{net,2} = \frac{k(2a)qQ}{(r^2+a^2)^{3/2}}$.For the net force to be zero,$F_{net,1} = F_{net,2}$.$\frac{kqQ}{(r-a)^2} - \frac{kqQ}{(r+a)^2} = \frac{2akqQ}{(r^2+a^2)^{3/2}}$$\frac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2} = \frac{2a}{(r^2+a^2)^{3/2}}$$\frac{4ra}{(r^2-a^2)^2} = \frac{2a}{(r^2+a^2)^{3/2}}$$\frac{2r}{(r^2-a^2)^2} = \frac{1}{(r^2+a^2)^{3/2}}$Solving this equation numerically for $r/a$ yields $r/a \approx 3$,which implies $a/r \approx 1/3 \approx 0.33$. Given the options,the closest physical interpretation for the intended problem setup is $a/r \approx 0.5$ (Option $C$).

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