Answer
(a) $E = 9.71 \times 10^{4} \mathrm{V} / \mathrm{m}$
(b) $q = 3.03 \times 10^{-11} \mathrm{C} $
Work Step by Step
(a) The electric field for the cylindrical shape is given by
\begin{align}
E&=\frac{V_{a b}}{\ln (b / a)} \frac{1}{r}\\
& =\frac{50\times 10^{3} \mathrm{V}}{\ln \left(0.14 \mathrm{m} / 90 \times 10^{-6} \mathrm{m}\right)(0.07004 \mathrm{m})}\\
& =\boxed{9.71 \times 10^{4} \mathrm{V} / \mathrm{m}}
\end{align}
(b) The force exerted is ten times the weight, so it is given by
$$F =10 mg = 10\left(30 \times 10^{-9} \mathrm{kg}\right)\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) =2940 \times 10^{-9} \,\text{N}$$
Hence, we could get the charge by
$$ q=\frac{F}{E}=\frac{2940 \times 10^{-9} \,\text{N} }{9.71 \times 10^{4} \mathrm{V} / \mathrm{m}}=\boxed{3.03 \times 10^{-11} \mathrm{C}} $$
