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(B) The electric field due to an infinite line charge at a distance $r$ is $E = \frac{\lambda}{2\pi \varepsilon_0 r}$.Since the electric field is cons

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$\frac{q\lambda}{4\pi \varepsilon_0} \ln 2$

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(B) The electric field due to an infinite line charge at a distance $r$ is $E = \frac{\lambda}{2\pi \varepsilon_0 r}$.Since the electric field is conservative,the work done in moving a charge $q$ depends only on the initial and final positions.Point $A$ is at a distance $r_A = a$ from the $x$-axis.Point $C$ is at a distance $r_C = \sqrt{a^2 + a^2} = a\sqrt{2}$ from the $x$-axis.The work done $W_{CA}$ is given by $W = q(V_A - V_C) = -q \int_{r_C}^{r_A} E dr$.$W = -q \int_{a\sqrt{2}}^{a} \frac{\lambda}{2\pi \varepsilon_0 r} dr = -\frac{q\lambda}{2\pi \varepsilon_0} [\ln r]_{a\sqrt{2}}^{a}$.$W = -\frac{q\lambda}{2\pi \varepsilon_0} (\ln a - \ln(a\sqrt{2})) = -\frac{q\lambda}{2\pi \varepsilon_0} \ln(\frac{a}{a\sqrt{2}})$.$W = -\frac{q\lambda}{2\pi \varepsilon_0} \ln(\frac{1}{\sqrt{2}}) = \frac{q\lambda}{2\pi \varepsilon_0} \ln(\sqrt{2}) = \frac{q\lambda}{2\pi \varepsilon_0} \ln(2^{1/2}) = \frac{q\lambda}{4\pi \varepsilon_0} \ln 2$.

For an infinite line of charge having charge density $\lambda$ lying along the $x$-axis,the work required in moving a charge $q$ from point $C$ to point $A$ along the arc $CA$ is:

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