Question

The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 4.7. (a) The potential on the outer surface of the membrane is 79.5 mV greater than that on the inside surface. How much charge resides on the outer surface

Answer 1

Answer:

Q = 1.2*10⁻¹² C

Explanation:

• For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:

       $C = \frac{Q}{V} (1)$

• For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

       $C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)$

• So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
• Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

       $Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)$