A capacitor holding 4.0 J of stored energy is connected to an identical capacitor that has no electric field between its plates. Determine the total energy stored across the two capacitors.

Step-by-step solution

Let the capacitance of each capacitor be C and the initial charge on the first capacitor be Q . Initial energy of the first capacitor is U_ i = Q^2 2C = 4.0 J . When it is connected to an identical uncharged capacitor, the total charge Q is conserved. The equivalent capacitance of the parallel combination becomes C_ eq = C + C = 2C . The final total energy stored in the system is U_ f = Q^2 2C_ eq = Q^2 2(2C) . U_ f = 1 2 ( Q^2 2C ) = 1 2 U_ i . Substituting the given value, U_ f = 1 2 4.0 J = 2.0 J .