a) (i) \(\displaystyle{V}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\left({\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}-{\ln{{\left({\frac{{{b}}}{{{b}}}}\right)}}}\right)}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}\)

(ii) \(\displaystyle{V}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\left({\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}-{\ln{{\left({\frac{{{b}}}{{{b}}}}\right)}}}\right)}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}\)

(iii) \(\displaystyle{V}={0}\)

b) \(\displaystyle{V}_{{{a}{b}}}={V}{\left({a}\right)}-{V}{\left({b}\right)}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}\)

c) Between the cylinders:

\(\displaystyle{V}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}={\frac{{{V}_{{{a}{b}}}}}{{{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}}}}{\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}\)

\(\displaystyle\therefore{E}={\frac{{{d}{V}}}{{{d}{r}}}}=-{\frac{{{V}_{{{a}{b}}}}}{{{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}}}}{\frac{{{d}}}{{{d}{r}}}}{\left({\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}\right)}={\frac{{{V}_{{{a}{b}}}}}{{{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}}}}{\frac{{{1}}}{{{r}}}}\)

d) The potential difference between the two cylinders is identical to that in part (b) even if the outer cylinder has no charge.

(ii) \(\displaystyle{V}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\left({\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}-{\ln{{\left({\frac{{{b}}}{{{b}}}}\right)}}}\right)}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}\)

(iii) \(\displaystyle{V}={0}\)

b) \(\displaystyle{V}_{{{a}{b}}}={V}{\left({a}\right)}-{V}{\left({b}\right)}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}\)

c) Between the cylinders:

\(\displaystyle{V}={\frac{{\lambda}}{{{2}\pi\epsilon_{{0}}}}}{\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}={\frac{{{V}_{{{a}{b}}}}}{{{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}}}}{\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}\)

\(\displaystyle\therefore{E}={\frac{{{d}{V}}}{{{d}{r}}}}=-{\frac{{{V}_{{{a}{b}}}}}{{{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}}}}{\frac{{{d}}}{{{d}{r}}}}{\left({\ln{{\left({\frac{{{b}}}{{{r}}}}\right)}}}\right)}={\frac{{{V}_{{{a}{b}}}}}{{{\ln{{\left({\frac{{{b}}}{{{a}}}}\right)}}}}}}{\frac{{{1}}}{{{r}}}}\)

d) The potential difference between the two cylinders is identical to that in part (b) even if the outer cylinder has no charge.