Answered: a) The potential on the outer surface…

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The membrane that surrounds a certain type of living cell has a surface area of 5.45 10^-9 m² and a thickness of 9.90 10^-9 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.50.

(a) The potential on the outer surface of the membrane is +65.4 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?

Expert Solution

Step 1

A capacitor is a system of two conductors carrying equal and opposite charges placed at a very short distance. For a capacitor, the charge on a conductor is directly proportional to the potential difference between the conductor.

$Q = C V$

$C$ is a constant and is known as the capacitance. The SI unit of capacitance is Farad $(F)$ . One of the most common capacitors is a parallel plate capacitor where two plates of equal dimension and carrying opposite charge density are placed at a very short distance.

Let the distance between the plates be $d$ and the surface area of the plates be $A$ . The electric field between the plates

$E = \frac{σ}{ε_{0} K}$

$K$ is the dielectric constant of the material between the plates. We know that the charge on the plates

$σ = \frac{Q}{A}$

This gives

$E = \frac{1}{ε_{0} K} (\frac{Q}{A})$

The potential difference between the plates

$V = E d$

Therefore the capacitance of the capacitor

$C = \frac{Q}{V} = \frac{Q}{E d} = \frac{Q ε_{0} K A}{Q d} = K ε_{0} \frac{A}{d}$

Step 2

In the given question we have the surface area of the cell $A = 5.45 \times 10^{- 9} m^{2}$

the thickness of the cell $d = 9.9 \times 10^{- 9} m$

The dielectric constant of the medium $K = 5.50$

Therefore the capacitance of the cell wall

$C = K ε_{0} \frac{A}{d} = 5.5 \times 8.85 \times 10^{- 12} \times \frac{5.45 \times 10^{- 9}}{9.9 \times 10^{- 9}} = 2.68 \times 10^{- 11} F$

a. The potential difference between the walls $V = 65.4 m V = 65.4 \times 10^{- 3} V$

Therefore the charge that resides on the surface is

$Q = C V = 2.68 \times 10^{- 11} \times 65.4 \times 10^{- 3} = 1.752 \times 10^{- 12} C$

b. Potassium ions carry $+ 1$ unit of charge. Let the number of potassium ions be N. Therefore by quantization of charge we have

$Q = N e$

This gives

$1.752 \times 10^{- 12} = N \times 1.6 \times 10^{- 19} \Rightarrow N = \frac{1.752 \times 10^{- 12}}{1.6 \times 10^{- 19}} = 1.0935 \times 10^{7}$

Step by step

Solved in 3 steps with 1 images

a) The potential on the outer surface of the membrane is +65.4 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?