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(B) The electric potential $V$ at a distance $r$ from an infinitely long line charge with linear charge density $\lambda$ is given by $V(r) = -\frac{\

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$(\lambda \ln 2) / (\pi \varepsilon_0)$

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(B) The electric potential $V$ at a distance $r$ from an infinitely long line charge with linear charge density $\lambda$ is given by $V(r) = -\frac{\lambda}{2\pi\epsilon_0} \ln(r) + C$.The work done by an external agent in moving a charge $q$ from point $A$ to point $B$ is $W_{ext} = q(V_B - V_A)$. Here,$q = 1$.The potential at any point $(x, y, z)$ due to the three wires is:$V(x, y, z) = V_x + V_y + V_z$$V(x, y, z) = -\frac{2\lambda}{2\pi\epsilon_0} \ln(\sqrt{y^2+z^2}) - \frac{3\lambda}{2\pi\epsilon_0} \ln(\sqrt{x^2+z^2}) - \frac{\lambda}{2\pi\epsilon_0} \ln(\sqrt{x^2+y^2})$At point $A(1, 1, 1)$:$V_A = -\frac{\lambda}{2\pi\epsilon_0} [2 \ln(\sqrt{2}) + 3 \ln(\sqrt{2}) + 1 \ln(\sqrt{2})] = -\frac{6\lambda}{2\pi\epsilon_0} \ln(\sqrt{2}) = -\frac{3\lambda}{2\pi\epsilon_0} \ln(2)$At point $B(0, 1, 1)$:$V_B = -\frac{2\lambda}{2\pi\epsilon_0} \ln(\sqrt{1^2+1^2}) - \frac{3\lambda}{2\pi\epsilon_0} \ln(\sqrt{0^2+1^2}) - \frac{\lambda}{2\pi\epsilon_0} \ln(\sqrt{0^2+1^2})$$V_B = -\frac{2\lambda}{2\pi\epsilon_0} \ln(\sqrt{2}) - 0 - 0 = -\frac{\lambda}{2\pi\epsilon_0} \ln(2)$Work done $W_{ext} = V_B - V_A = -\frac{\lambda}{2\pi\epsilon_0} \ln(2) - (-\frac{3\lambda}{2\pi\epsilon_0} \ln(2)) = \frac{2\lambda}{2\pi\epsilon_0} \ln(2) = \frac{\lambda \ln 2}{\pi \epsilon_0}$.

The diagram shows three infinitely long uniform line charges placed on the $X, Y$ and $Z$ axes with linear charge densities $2\lambda, 3\lambda$ and $\lambda$ respectively. The work done by an external agent in moving a unit positive charge from $(1, 1, 1)$ to $(0, 1, 1)$ is equal to:

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