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(A) The electric field due to an infinite wire at a distance $r$ is $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.Given $\lambda = 4 \pi \varepsilon_0$,

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work done by external agent $= 4 \, \ln 2$

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(A) The electric field due to an infinite wire at a distance $r$ is $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.Given $\lambda = 4 \pi \varepsilon_0$,we have $E = \frac{4 \pi \varepsilon_0}{2 \pi \varepsilon_0 r} = \frac{2}{r}$.The potential difference between points $A$ $(r_A = 2 \, m)$ and $B$ $(r_B = 1 \, m)$ is:$V_B - V_A = - \int_{r_A}^{r_B} E \, dr = - \int_{2}^{1} \frac{2}{r} \, dr = -2 [\ln r]_2^1 = -2 (\ln 1 - \ln 2) = 2 \ln 2$.Since the particle moves with constant speed,the net work done is zero $(W_{ext} + W_{elec} = 0)$.The work done by the electric force is $W_{elec} = q(V_A - V_B) = q(-(V_B - V_A)) = 2 	imes (-2 \ln 2) = -4 \ln 2$.The work done by the external agent is $W_{ext} = -W_{elec} = 4 \ln 2$.

The figure shows a positively charged infinite wire. $A$ particle of charge $q = 2 \, C$ moves from point $A$ to $B$ with constant speed. (Given linear charge density on the wire is $\lambda = 4 \pi \varepsilon_0$)

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