Question

Suppose the initial charge on the capacitor in Exercise 7 is 6mC. What is the total energy at later time ?

Answer 1

Given, charge on the capacitor
`Q=6mC = 6xx10^(-3)C`
`C=30mu F=30xx10^(-6)F`
Energy stored in the circuit
`E =(Q^(2))/(2C)=((6xx10^(-3))^(2))/(2xx30xx10^(-2))=(36)/(60)=0.6J`
After some time, the energy is shared between C and L, but the total energy remains constant. So, we assume that there is no loss of energy.