Answer to Question #132393 in Mechanics | Relativity for Jessica

Question

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10^-9 m2 and a thickness of 1.5 x 10^-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.3. (a) The potential on the outer surface of the membrane is +89.0 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

Concept

The capacitance of cell membrane is given by the expression:

$$c= \dfrac{k\epsilon_{o}A }{d}$$

The charge on the capacitor is given by:

$$q= CV$$

We have to charge on the outer surface of the membrane and also the number of ions present on it.

The Capacitance

$$C= \dfrac{k\epsilon_{o}A }{d}$$

$$C= \dfrac{(5.3)\cdot(8.85*10^{-12})\cdot(5.3*10^{-9})}{1.5*10^{-8}}$$

$$C=1.65*10^{-11}F$$

Part(a):The charge on capacitor

$$q=CV$$

$$q=(1.65*10^{-11}F)(89.0*10^{-3}V)$$

$$q=1.46*10^{-12}C$$

Part(b): Number of Cations

From the law of quantization, we can find the number of cations:

$$N= \dfrac{q}{e}$$

$$N= \dfrac{1.46*10^{-12}C}{1.6*10^{-19}C}$$

$$N=9.12*10^{6}$$

Final answer:

The charge is $1.46*10^{-12}C$ and number of cations are $9.12*10^{6}$