Question

An air-filled capacitor consists of twoparallel plates, each with an area of 7.60 cm2,separated by a distance of 1.90 mm.A 16.0 V potential difference isapplied to these plates. Calculate the electric field between theplates, the surface chare density, the capacitance and the chargeon each plates.

Answer 1

(A)8,421.1 V/m

(B)
`3.54 x 10^(-12) F`

(C)
`53.1 x 10^(-12) C`

(D)
`6.987 x 10^(-8) C/m^(2)`

Step-by-step explanation:

area (A) = 7.6 cm^{2} = 0.00076 m^{2}

distance (d) = 1.9 mm = 0.0019 m

potential difference (v) = 16 V

(A) electric field (E) = Δv / d = 16 / 0.0019 = 8,421.1 V/m

(B) capacitance = (∈₀A)/d

where ∈₀ = permitivity of free space =
`8.85 x 10^(-12)m^(-3).kg^(-1) .s^(4).A^(2)`

capacitance =
`(8.85 x 10^(-12)x0.00076)/(0.0019)`

capacitance (C) =
`3.54 x 10^(-12) F`

(C) charge (q) = C x v =
`3.54 x 10^(-12) x 16` =
`56.64x 10^(-12) C`

(D) surface charge density = charge (q) / area =
`56.64 x 10^(-12)` / 0.00076 =
`7.453 x 10^(-8) C/m^(2)`