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

Answer:

(A)8,421.1 V/m

(B) $3.54 x 10^{-12} F$

(C)  $53.1 x 10^{-12} C$

(D)  $6.987 x 10^{-8} C/m^{2}$

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 = $\frac{8.85 x 10^{-12}x0.00076}{0.0019}$

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  =  $7.453 x 10^{-8} C/m^{2}$