Question

The layer that surrounds a certain type of living cell has a surface area of 4.5 x 10^-9 m² and a thickness of 1.4 x 10^-8 m. Assume that the layer behaves like a parallel plate capacitor and has a dielectric constant of 5.1. (a) The potential on the outer surface of the layer is +53.4 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?

Answer 1

PART A

We know that the capacitance C, of a parallel plate capacitor with surface area A , distance between plates d and a dielectric between the plates with a dielectric constant €_r is given as

C=€₀€_rA/d , where€₀ is the permittivity of free space=8.854×10^-12

Now we are given in the question d,A and dielectric constant so we can calculate the capacitance as

We have calculated the capacitance, we know that

Where Q is the amount of charge on each plate ( -Q on one and +Q on another)and V is the potential difference between the plates.

V is given to us as,

Substituting in the above equation of Q we get

Therefore,

Approximately Q=7.599*10^-13 C charge resides on the surface.

PART B

To calculate the number of ions that produces the total charge we simply divide the total charge by the amount of charge on one ion which is e=1.6*10^-19C

Number of ions So 4.749×10⁶ K⁺ ions are present on the outer surface.

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