Two electric dipoles A and B with respective | vedclass

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(A) The potential due to an electric dipole at a point on its axis is given by $V = \frac{kp \cos 	heta}{r^2}$. Since the point lies on the axis of b

Vedclass · Accepted Answer

$\frac{{\sqrt 2 R}}{{\sqrt 2 + 1}}$

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(A) The potential due to an electric dipole at a point on its axis is given by $V = \frac{kp \cos 	heta}{r^2}$. Since the point lies on the axis of both dipoles,$	heta = 0^\circ$ or $180^\circ$.Let the point be at a distance $x$ from dipole $A$. Then the distance from dipole $B$ is $(R - x)$.The potential due to dipole $A$ at this point is $V_A = \frac{k(4qa)}{x^2}$ (taking magnitude).The potential due to dipole $B$ at this point is $V_B = \frac{k(2qa)}{(R - x)^2}$.Equating the potentials,we get:$\frac{k(4qa)}{x^2} = \frac{k(2qa)}{(R - x)^2}$$\frac{2}{x^2} = \frac{1}{(R - x)^2}$Taking the square root on both sides:$\frac{\sqrt{2}}{x} = \frac{1}{R - x}$$\sqrt{2}(R - x) = x$$\sqrt{2}R - \sqrt{2}x = x$$\sqrt{2}R = x(1 + \sqrt{2})$$x = \frac{\sqrt{2}R}{\sqrt{2} + 1}$

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