A parallel plate capacitor of capacitance \(40\text{ }\mu F\) is connected to a \(100\text{ }\text{V}\) power supply. Th…

1. Initial Charge Stored:
   The charge stored in a capacitor is given by:

   \(Q=CV\)

   Substituting \(C=40\text{ }\mu \text{F}\) and \(V=100\text{ }\text{V}\):

   \(Q=40\times 1{0}^{−6}\times 100=4\text{ }\text{mC}\mathrm{.}\)

2. Capacitance with Dielectric:
   When the dielectric is introduced, the capacitance becomes:

   \({C}^{′}=K⋅C=2⋅40=80\text{ }\mu \text{F}\mathrm{.}\)

3. Charge with Dielectric:

   \({Q}^{′}={C}^{′}⋅V=80\times 1{0}^{−6}⋅100=8\text{ }\text{mC}\mathrm{.}\)

   The extra charge stored is:

   \(\Delta Q={Q}^{′}−Q=8−4=4\text{ }\text{mC}\mathrm{.}\)

4. Change in Electrostatic Energy:
   The energy stored in a capacitor is given by:

   \(U=\frac{1}{2}C{V}^{2}\mathrm{.}\)
   • Initial energy: \(U=\frac{1}{2}⋅40\times 1{0}^{−6}⋅(100{)}^{2}=0.2\text{ }\text{J}\mathrm{.}\)
   • Final energy with dielectric: \({U}^{′}=\frac{1}{2}⋅80\times 1{0}^{−6}⋅(100{)}^{2}=0.4\text{ }\text{J}\mathrm{.}\)

   The change in energy is:

   \(\Delta U={U}^{′}−U=0.4−0.2=0.2\text{ }\text{J}\mathrm{.}\)

Thus, the extra charge stored is \(4\text{ }\text{mC}\) and the change in energy is \(0.2\text{ }\text{J}\).