A large parallel plate capacitor has plate separation 1.00 cm and plate area 314

Question

A large parallel plate capacitor has plate separation 1.00 cm and plate area 314 cm2 with air between the plates. The capacitor is connected to a 20.0 V battery. With the battery still connected, a slab of strontium titanate is inserted so that it completely fills the gap between the plates. (a) Find the charge on the plates, the electric field between the plates, and the energy stored in the capacitor before the slab is inserted. (b) Find the charge on the plates, the electric field, the potential difference, and the energy stored in the capacitor after the slab is inserted.

Explanation / Answer

C= K*Eo*A/D

a) charge stored = CV

                             = Eo*A/D*V

                               = 8.854x10-12*0.0314 *20/0.01

                               = 5.56*10-10 C

   Electric field between plates = V/d

                                                  = 20/0.01

                                                   = 2000 N/C

Energy stored in capacitor = 1/2*Q*V

                                                        = 0.5*5.56*10-10*20

                                                = 5.56*10-9 J

b) with strontium titanate charge on plates = 310*5.56*10-10

                                                                                     = 1.723*10-7 C

       Potential difference between plates = 20 V

       Electric field =2000 N/C

energy stored = 0.5*1.723*10-7*20

                        =   1.723 *10-6 J

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