The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination?

Answer:net voltage will become half that of initial voltage nowExplanation:Initially two capacitors are connected across a given voltage differenceSo the net charge on the two capacitors is given as[tex]Q = C_{eq} V[/tex][tex]Q = \frac{C}{2} V[/tex]now when these are connected in parallel with no change in chargethen total charge on each plate is given[tex]Q = \frac{CV}{2} + \frac{CV}{2}[/tex][tex]Q = CV[/tex]also we know that[tex]C_{eq} = 2C[/tex]now we will have[tex]V_{net} = \frac{Q_{net}}{C_{eq}}[/tex][tex]V_{net} = \frac{CV}{2C} = \frac{V}{2}[/tex]so net voltage will become half that of initial voltage now