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(D) Consider the arrangement given.The electric field $E_{1}$ produced by the first infinitely long conductor at a perpendicular distance $R$ is given

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$r^{0}$

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(D) Consider the arrangement given.The electric field $E_{1}$ produced by the first infinitely long conductor at a perpendicular distance $R$ is given by $E_{1} = \frac{2k\lambda_{1}}{R}$.Consider a small element of length $dl$ on the second conductor at a distance $l$ from the point closest to the first conductor. The distance $R$ from the first conductor to this element is $R = \sqrt{r^{2} + l^{2}}$.The force $dF$ on the charge element $dq = \lambda_{2} dl$ is $dF = E_{1} dq = \frac{2k\lambda_{1}}{\sqrt{r^{2} + l^{2}}} \lambda_{2} dl$.Due to symmetry,the components of force perpendicular to the second conductor cancel out. The net force $F$ is the integral of the components along the second conductor: $F = \int_{-\infty}^{\infty} dF \cos 	heta$,where $\cos 	heta = \frac{r}{R} = \frac{r}{\sqrt{r^{2} + l^{2}}}$.Substituting $\cos 	heta$,we get $F = \int_{-\infty}^{\infty} \frac{2k\lambda_{1}\lambda_{2}}{\sqrt{r^{2} + l^{2}}} \cdot \frac{r}{\sqrt{r^{2} + l^{2}}} dl = 2k\lambda_{1}\lambda_{2}r \int_{-\infty}^{\infty} \frac{dl}{r^{2} + l^{2}}$.Using the substitution $l = r 	an 	heta$,$dl = r \sec^{2} 	heta d	heta$,the integral becomes $2k\lambda_{1}\lambda_{2}r \int_{-\pi/2}^{\pi/2} \frac{r \sec^{2} 	heta d	heta}{r^{2} \sec^{2} 	heta} = 2k\lambda_{1}\lambda_{2} \int_{-\pi/2}^{\pi/2} d	heta = 2k\lambda_{1}\lambda_{2} [	heta]_{-\pi/2}^{\pi/2} = 2\pi k\lambda_{1}\lambda_{2}$.Since the result is independent of $r$,the force is proportional to $r^{0}$.

Two mutually perpendicular infinitely long straight conductors carrying uniformly distributed charges of linear densities $\lambda_{1}$ and $\lambda_{2}$ are positioned at a distance $r$ from each other. Force between the conductors depends on $r$ as

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