The membrane that surrounds a certain type of living cell has a surface area of 4.4 x 10-9 m2 and a thickness of 1.6 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 4.1. (a) The potential on the outer surface of the membrane is +66.4 mV greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to K+ ions (charge +e), how many such ions are present on the outer surface?

C=q/ Δφ,

C=εε₀A/d,
q/ Δφ = εε₀A/d,
q= εε₀A Δφ /d,
where ε =4.1, ε₀=8.85 •10⁻¹² F/m,
A=4.4•10⁻⁹m²,
d=1.64.4•10⁻⁸ m,

N=q/e=q/1.6•10⁻¹⁹ = …

To solve this problem, we need to use the formula for capacitance of a parallel plate capacitor:

C = (ε₀ * ε * A) / d

Where:
C = Capacitance
ε₀ = Permittivity of free space (8.85 x 10^-12 F/m)
ε = Dielectric constant
A = Surface area of the membrane
d = Thickness of the membrane

(a) To find the charge on the outer surface, we can use the formula for capacitance:

C = Q / V

Where:
Q = Charge on the outer surface
V = Potential difference between the outer and inner surfaces

Rearranging the formula, we get:

Q = C * V

Substituting the given values:

C = (ε₀ * ε * A) / d = (8.85 x 10^-12 F/m * 4.1 * 4.4 x 10^-9 m^2) / (1.6 x 10^-8 m)

V = 66.4 mV = 66.4 x 10^-3 V

Calculating the charge:

Q = (8.85 x 10^-12 F/m * 4.1 * 4.4 x 10^-9 m^2) / (1.6 x 10^-8 m) * (66.4 x 10^-3 V)

(b) To find the number of K+ ions on the outer surface, we need to know the charge of each K+ ion. Given that the charge of a K+ ion is +e, we can find the number of ions:

Number of ions = Q / e

Substituting the values:

Number of ions = Q / +e

Now you can calculate both the charge on the outer surface and the number of K+ ions present on the outer surface using the given information and the equations provided above.