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(D) The electric field $E$ on the axis of a uniformly charged ring of radius $R$ and charge $Q$ at a distance $x$ from the center is given by $E = \fr

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zero

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(D) The electric field $E$ on the axis of a uniformly charged ring of radius $R$ and charge $Q$ at a distance $x$ from the center is given by $E = \frac{kQx}{(R^2 + x^2)^{3/2}}$.Given $x = R/\sqrt{2}$,the electric field is $E = \frac{kQ(R/\sqrt{2})}{(R^2 + R^2/2)^{3/2}} = \frac{kQR/\sqrt{2}}{(3R^2/2)^{3/2}} = \frac{kQR/\sqrt{2}}{3\sqrt{3}R^3/2\sqrt{2}} = \frac{2kQ}{3\sqrt{3}R^2}$.The force on a dipole $P$ in an electric field is $F = P \frac{dE}{dx}$.We need to find the derivative $\frac{dE}{dx}$ at $x = R/\sqrt{2}$.$\frac{dE}{dx} = kQ \left[ \frac{(R^2+x^2)^{3/2} - x \cdot \frac{3}{2}(R^2+x^2)^{1/2} \cdot 2x}{(R^2+x^2)^3} \right] = kQ \left[ \frac{R^2+x^2 - 3x^2}{(R^2+x^2)^{5/2}} \right] = kQ \frac{R^2-2x^2}{(R^2+x^2)^{5/2}}$.At $x = R/\sqrt{2}$,$R^2 - 2x^2 = R^2 - 2(R^2/2) = 0$.Therefore,the force $F = P \cdot 0 = 0$.

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