Question

A parallel-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?

Answer 1

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

Step-by-step 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 =
`((8.85* 10^(-12))(0.2))/(0.007)`

= 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

=
`((140))/(0.007)`

= 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 =
`((8.85* 10^(-12))(0.2))/(0.016)`

=
`110.6* 10^-12`

= 110.6 pF

B) Charge = Q =
`350* 10^-12`

C) E= 20,000V

D) Energy =
`0.5 (Q^2)/(C)`

=
`0.5 ((350* 10^-12)^2)/((110.6* 10^-12))`

=
`5.54* 10^(-10)` J