Question

The membrane that surrounds a certain type of living cell has a surface area of 7.1 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 4.4. (a) The potential on the outer surface of the membrane is 75.9 mV greater than that on the inside surface. How much charge resides on the outer surface

Answer 1

Answer:

Q = +1.4 pC

Explanation:

• The capacitance of any capacitor, by definition, is as follows:
•  $C = \frac{Q}{V} (1)$
• Applying Gauss'Law and the definition of electric potential, it can be showed that the capacitance of a parallel-plate capacitor can be expressed as follows:
•  $C = \frac{\epsilon_{r}*\epsilon_{0} * A}{d} (2)$
• Where εr is the dielectric constant of the material that fills the space between the plates, A is the area of one of the plates, and d, is the separation between them.
• Replacing by the givens in (2) we can find the value of the capacitance C, as follows:

       $C = \frac{\ 4.4*8.85e-12C2/N*m2*7.1e-9m2}{1.5e-8m} = 18.4e-12 F = 18.4 pF$

• Replacing the values of C and V in (1), we can solve for Q, as follows:

       $Q = C*V = 18.4e-12 F* 75.9e-3 V = 1.4e-12 C = +1.4 pC$

• As the outer surface is at a higher potential that the inside surface, the charge on it must be positive, and is equal to +1.4 pC.