Mutual capacitance

E_p= λ /2∩εx + λ/2∩ε(b-x)
THUS, W =work done to take test charge from positive to negative surface =
∫{from (a) to (b-a)} E_p dx
=..................
=......λ /2∩ε{2 log (b-a) -2log a}
= λ /2∩ε( ln b/a) =V
thus, C per unit length = λ/V = ∩ε / (ln b/a)

yeah but see what answer irodov has givne

C = \pi \epsilon _{0}ln(b/a)

how do u get that?

look at the next answer 3.109...its definite that there is a printing mistake in 3.108 ...the / sign has got omitted by mistake !