Aparallel-plate air capacitor is made from two plates 0.200 msquare, spaced 0.700 cm apart. it is connected to a 140 v battery. part a: what is the capacitance?
part b: what is the charge on each plate?
part c: what is the electric field between the plates?
part d: what is the energy stored in the capacitor?
part e: if the battery is disconnected and then the plates are pulled apart to a separation of 1.60 cm, what are the answers to parts a, b, c, and d?

A)2.53 × 10⁻¹² F

B) 350× 10⁻¹² C

C) 20,000V/m

D) 2.48 × 10⁻⁸J

E) 110.6 pF , 350 pC, 20,000V/m, 5.54 x 10⁻¹⁰ J

Explanation:

A) Given:

Area of the plates = A = 0.200 m²

Distance of separation of the plates = d = 0.007 m

Battery voltage = V = 140 V

Formula for Capacitance : C = ε₀ A/d

ε₀ = Permittivity of free space =  8.85×10⁻¹²    SI units

C =

   =  2.53 × 10⁻¹² F    

B) Charge on each plate: q = C V= (2.53 × 10⁻¹²)(140)    

                                                       = 350 × 10⁻¹² C

C) Electric field between the plates = E = V/d

                                                                  =

                                                                    = 20000 V/m

D) Energy = 0.5 C V²

                = (0.5)(2.53 × 10⁻¹²)(140)²

                = 2.48 × 10⁻⁸ J

E)  When the battery is disconnected, the charge and electric field remain the same.

A) Since d = 0.0016 m now,

C =

   =

    = 110.6 pF

B) Charge = Q =

C) E= 20,000V

D) Energy =

                  =

                  = J

im guessing the west of station b