Answer
45.2 mA
Work Step by Step
We know that Maximum $U_{E}$= Maximum $U_{B}$= Total energy.
$\implies \frac{q_{max}^{2}}{2C}=\frac{Li_{max}^{2}}{2}$
$\implies i_{max}=\sqrt {\frac{q_{max}^{2}}{LC}}=\frac{q_{max}}{\sqrt {LC}}$
Given: $q_{max}= 3.00\,\mu C$, L=1.10 mH, C= $4.00\,\mu F$
Result: $ i_{max}=\frac{3.00\times10^{-6}C}{\sqrt {1.10\times10^{-3}H\times4.00\times10^{-6}F}}=45.2\times10^{-3}A=45.2\,mA$
