Lc circuit consists of a 20.0 mh inductor and a 0.400 µf capacitor. if the maximum instantaneous current is 0.500 a, what is the greatest potential difference across the capacitor

The governing equation for this situation is the energies in both the inductor and the capacitor. These two energies must be equal and are shown by the equation below:

LI^2/2 = CV^2/2,

Where

L = 20.0 mH = 20*10^-3 H

I = 0.5 Amps

C=0.4 uf = 0.4*10^-6 F

Substituting;

(20*10^-3*0.5^2) / 2 = (0.4*10^-6*V^2) / 2

0.0025 = 2*01^-7V^2

V^2 = 0.0025 / (2*10^-7) = 12500

V = Sqrt (12500) = 111.8 V

The greatest potential across the capacitor is 111.8 V.