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(B) The electric field due to an infinite line charge at a distance $r$ is given by $E_{line} = \frac{2k\lambda}{r}$.The electric field due to a point

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(B) The electric field due to an infinite line charge at a distance $r$ is given by $E_{line} = \frac{2k\lambda}{r}$.The electric field due to a point charge $q$ at a distance $(d-r)$ from it is $E_{point} = \frac{kq}{(d-r)^2}$.For the total electric field to be zero,the magnitudes must be equal: $E_{line} = E_{point}$.Substituting the values $\lambda = 1 \ C \ m^{-1}$,$q = 1 \ C$,and $d = 3 \ m$:$\frac{2k(1)}{r} = \frac{k(1)}{(3-r)^2}$$\frac{2}{r} = \frac{1}{(3-r)^2}$$2(3-r)^2 = r$$2(9 - 6r + r^2) = r$$18 - 12r + 2r^2 = r$$2r^2 - 13r + 18 = 0$Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$r = \frac{13 \pm \sqrt{169 - 144}}{4} = \frac{13 \pm 5}{4}$$r_1 = \frac{18}{4} = 4.5 \ m$ and $r_2 = \frac{8}{4} = 2 \ m$.Since the point must be between the origin and the charge $(0 < r < 3)$,the valid distance is $r = 2 \ m$.

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