The membrane that surrounds a certain type of living cell has a surface area of 4.9 x 10-9 m2 and a thickness of 1.2 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.5. (a) The potential on the outer surface of the membrane is +61.4 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?
To find the charge residing on the outer surface of the membrane, we can use the formula for capacitance of a parallel plate capacitor:
C = (ε0 * εr * A) / d
Where:
C is the capacitance,
ε0 is the permittivity of free space (8.85 x 10^-12 F/m),
εr is the relative permittivity (dielectric constant) of the membrane (5.5),
A is the surface area of the membrane (4.9 x 10^-9 m^2),
d is the thickness of the membrane (1.2 x 10^-8 m).
We can rearrange the formula to solve for the capacitance:
C = (ε0 * εr * A) / d
C = (8.85 x 10^-12 F/m * 5.5 * 4.9 x 10^-9 m^2) / (1.2 x 10^-8 m)
C = 2.139 x 10^-14 F
Now, the charge on a capacitor can be calculated using the formula:
Q = C * V
Where:
Q is the charge on the capacitor,
C is the capacitance (2.139 x 10^-14 F),
V is the potential difference between the plates (61.4 mV).
Converting the potential difference from millivolts to volts:
V = 61.4 mV * (1 V / 1000 mV)
V = 0.0614 V
Finally, we can calculate the charge on the outer surface:
Q = C * V
Q = 2.139 x 10^-14 F * 0.0614 V
Q = 1.312 x 10^-15 C
Therefore, the charge residing on the outer surface of the membrane is approximately 1.312 x 10^-15 Coulombs.
To find the number of K+ ions (charge +e) present on the outer surface, we need to know the charge on each K+ ion. Since the charge on one electron (e) is -1.6 x 10^-19 C, the charge on each K+ ion would be +1.6 x 10^-19 C.
Now, we can determine the number of K+ ions using the formula:
Number of K+ ions = (Charge on outer surface) / (Charge on each K+ ion)
Number of K+ ions = (1.312 x 10^-15 C) / (1.6 x 10^-19 C)
Calculating the division:
Number of K+ ions = 8.2 x 10^3 ions
Therefore, approximately 8.2 x 10^3 K+ ions are present on the outer surface of the membrane.