Answer:
(A)8,421.1 V/m
(B) [tex]3.54 x 10^{-12} F[/tex]
(C) [tex]53.1 x 10^{-12} C[/tex]
(D) [tex]6.987 x 10^{-8} C/m^{2}[/tex]
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 = [tex]8.85 x 10^{-12}m^{-3}.kg^{-1} .s^{4}.A^{2}[/tex]
capacitance = [tex]\frac{8.85 x 10^{-12}x0.00076}{0.0019}[/tex]
capacitance (C) = [tex]3.54 x 10^{-12} F[/tex]
(C) charge (q) = C x v = [tex]3.54 x 10^{-12} x 16[/tex] = [tex]56.64x 10^{-12} C[/tex]
(D) surface charge density = charge (q) / area = [tex]56.64 x 10^{-12} [/tex] / 0.00076 = [tex]7.453 x 10^{-8} C/m^{2}[/tex]
