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Question

(C) The electric field $E$ at a distance $r$ from an infinitely long charged wire is given by $E = \frac{2K\lambda}{r}$.The work done by an external a

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$-2K\lambda q \ln(2)$

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(C) The electric field $E$ at a distance $r$ from an infinitely long charged wire is given by $E = \frac{2K\lambda}{r}$.The work done by an external agent to move a charge $q$ is $W_{ext} = \Delta U = -\int_{a}^{2a} \vec{F} \cdot d\vec{r} = -\int_{a}^{2a} q\vec{E} \cdot d\vec{r}$.Substituting the expression for $E$: $W_{ext} = -\int_{a}^{2a} q \left( \frac{2K\lambda}{r} \right) dr$.$W_{ext} = -2K\lambda q \int_{a}^{2a} \frac{1}{r} dr$.$W_{ext} = -2K\lambda q [\ln(r)]_{a}^{2a}$.$W_{ext} = -2K\lambda q (\ln(2a) - \ln(a)) = -2K\lambda q \ln\left(\frac{2a}{a}\right)$.$W_{ext} = -2K\lambda q \ln(2)$.

An infinitely long uniformly charged string is placed along the $z-$axis. Its linear charge density is $\lambda$. $A$ point charge $q$ is moved from position $(a, 0, 0)$ to $(2a, 0, 0)$. The work done by the external agent is:

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