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 1.1 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.9. (a) The potential on the outer surface of the membrane is 85.9 mV greater than that on the inside surface. How much charge resides on the outer surface?

Answer 1

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane=$5.3\times 10^{-9} m^2$

Thickness of membrane=$1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor=$\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V=$85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface=$2.1\times 10^{-12} c$

Answer 2

Answer:

The charge on the outer surface is $2.15\times 10^{- 12}\ C$

Solution:

As per the question:

Surface Area, A = $5.3\times 10^{- 9}\ m^{2}$

Thickness, t = $1.1\times 10^{- 8}\ m$

Dielectric constant, K = 5.9

Potential on the on the membrane's outer surface, V = 85.9 mV = 0.0859 V

Now,

(a) To calculate the charge on the surface, Q:

We know that the capacitance of a parallel plate capacitor can be given as:

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

$C = \frac{5.9\times 8.85\times 10^{- 12}\times 5.3\times 10^{- 9}}{1.1\times 10^{- 8}} = 2.5\times 10^{- 11}\ F$

$Q = CV = 2.5\times 10^{- 11}\times 0.0859 = 2.15\times 10^{- 12}\ C$