A parallel-plate air capacitor of area A = 15.0 cm2 and plate separation d = 3.00 mm is charged by a battery to a voltage 58.0 V. If a dielectric material with ? = 4.60 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?

The additional charge that will flow from the battery onto the positive plate is 0.924 nC

Explanation:

Given;

Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²

Separation distance, d = 3.00 mm = 3 x 10⁻³ m²

voltage of the capacitor, V = 58.0 V

dielectric constant, k = 4.60

Initial Capacitance of the capacitor before the addition of dielectric material:

Initial charge across the parallel plates:

Q₁ = CV

Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C

Capacitance of the capacitor after the addition of dielectric material:

Cequ. = Ck

Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F

Final charge across the parallel plates:

Q₂ = Cequ. x V

Q₂ =  20.355 x 10⁻¹² x 58 = 11.806 x 10⁻¹⁰ C

Additional charge = Q₂ - Q₁

                              = 11.806 x 10⁻¹⁰ C - 2.567 x 10⁻¹⁰ C

                              = 9.239 x 10⁻¹⁰ C

                              = 0.924 nC

Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC

i'm on my way to history not a mistory.