A membrane that surrounds a cetain type of living cell has a surface area of?

Question

I am so confused with this problem. I do not even know where to begin. Can somebody help me? Thank you for your time.

A membrane that surrounds a cetain type of living cell has a surface area of 5.0e-9 m^2 and a thickness of 1e-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.

(a) The potential on the outer surface of the membrane is +60mV 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

The capacitance of the membrane is:

C = κƐ₀A / d

= (5 x (8.8 x 10^-12 F/m) x (5.00 x 10^-9 m²)) / (1.00 x 10^-8 m)

= 2.2 x 10^-11 F

(a) The charge is:

Q = C∆V

= (2.2 x 10^-11 F) x (0.06 V)

= 1.3 x 10^-12 C

(b) The number of potassium ions is:

# = q / e

= (1.3 x 10^-12 C) / (1.6 x 10^-19 C)

= 8.1 x 10^6 K+ ions

Answer 2

Capacitance is given by the equation:  C = epsilon_0*A/d

Using the thickness of the membrane (1e-8 m), you can determine the distance between the outer face and inner face of the membrane.  That's our variable "d".

We are given the surface area of the outer membrane, and since we're modeling the cell membrane as a parallel plate capacitor, we can assume the inner membrane has a similar surface area.  That's our "A"

Plug those variables into your equation for capacitance.  

I get 4.43*10^-12 F

You should multiply your answer by the dielectric constant (kappa = 5) to get your total capacitance.  

So that's now 2.21*10^-11 F

The first part of the question asks for charge.  Remember the equation Q = CV.  We just found our C, and we are given the change in voltage (0.060V), which is our V.  Solve for Q and you're done.  Remember that the |Q| is the same on both sides of the membrane (one side positive, one side negative).  

We finally get 1.33*10^-12 C

The second part asks you to find the amount of potassium ions on the outer surface.  Just take the Q value that you solved for above, divide it by "+e" and you should have your number of ions.  (e is just 1.60*10^-19 C per ion)

That's 8.3*10^6 ions.