Explanation
$$\textbf{Step 1: Uniform Charge distribution on outer surface [Refer Figure]}$$
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}+\dfrac {1}{b}\right )$$
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}-\dfrac {1}{b}\right )$$
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}-\dfrac {1}{b^2}\right )$$
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}+\dfrac {1}{b^2}\right )$$
We know that work done $$dW=qdV$$
here , $$\displaystyle W_{B\rightarrow A}=q{V_A-V_B}=q\left[\dfrac{Q}{4\pi\epsilon_0 (OA)}-\dfrac{Q}{4\pi\epsilon_0(OB)}\right]=\dfrac{qQ}{4\pi\epsilon_0}[\dfrac{1}{a}-\dfrac{1}{b}]$$ where $$OA=a, OB=b$$
From the expression of energy stored in capacitor,
$$U=\dfrac {1}{2}\dfrac {q^2}{C}$$
$$\therefore [C] = \left [\dfrac {q^2}{U} = \right ]\left [\dfrac {A^2T^2}{ML^2T{^-2}}\right ]$$
$$=[M^{-1}L^{-2}T^4A^2]$$
Work done by field $$W=-\Delta U={U}_{A}-{U}_{B}=(-e){V}_{A}-(-e){V}_{B}$$
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