Question

The membrane that surrounds a certain type of living cell has a surface area of 4.3 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 4.2. (a) The potential on the outer surface of the membrane is 85.7 mV greater than that on the inside surface. How much charge resides on the outer surface

Answer 1

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell  $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate  $(d) = 1.1 \times 10^{-8} m$  

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

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

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

      $Q = \frac{k \epsilon A \Delta V }{d}$

      $Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

      $Q = 1.245 \times 10^{-12}$