Question

Two identical capacitors are connected in parallel, and each acquires a charge Q0 = 1.12E-6 C when connected to a source of voltage V0 = 2.45 V. The voltage source is disconnected, and a dielectric with constant ? = 2.00 is inserted to fill the space between the plates of one of the capacitors. What is the smaller of the two final capacitor charges? What is the larger of the two final charges? What is the final volatge across the capacitors?

Answer 1

Thwe total charge on the two capacitors is Q = 2Q_o

The capacitance without the dielectric is C

The potential difference across eachcapacitor is V₀

The charge on each capacitor is

          Q_o = C.V_o      ----------------(1)

Total charge when the battery is disconnected is

               Q = 2Q_o

_{The capacitance of capacitor when the dielectric is introduced in one capacitor is}

_{C' = k.C}

_{The equivalent capacitance for parallel combination is}

_{C" = C+kC}

        _{= C (k+1)}

_{The potential difference across each capacitance is samem for the capacitors are still connected in parallel,}

_{Let us consider the new common potential difference = V'}

      _{V' = 2Qo/(k+1)C}

           = 2V_o/(k+1)           -------(2)

The charge on first capacitor is

     Q₁ = C.V'

          = C.2Q_o/(k+1)C

          = 2Q_o/(k+1)

          = 2(1.12x10^-6C)/(2.00+1)

          =0.75x10^-6C

The Charge on C₂( With dielectric) is,

        Q₂ = C'.V'

             = kC.2Q_o/(k+1)C

             = 2kQ_o/(k+1)

             = 2(2.00)(1.12x10^-6C)/(2.00+1)   

             = 1.49x10^-6 C

The new common potential difference is

          V' = Q₁/C

              = (2Q_o/(k+1))/(C)

             = (2Q_o/((2.00)+1))/(C)

              = 0.67V₀   

                 = (0.67)(2.45 V)

             =1.63V