A simplified model for an erythrocyte is a spherical capacitor with a positive inner surface of area A and a membrane of thickness b. Use potential difference of 94 mV across the membrane and a dielectric constant K = 5.0. If the capacitance of the membrane is C = 0.15 pF, what is the radius of the erythrocyte? The thickness of the membrane is b = 90 nm.
The way I thought to solve for this question was that I found the area using the Capacitance of a dielectric eqn, then from there used A = pi*r^2 to solve for r...is this the correct approach? I'm just concerned because I'm not using the potential difference anywhere...
The membrane surface area should be
A = 4 pi r^2
for a spherical cell
Thanks, so there wouldn't be any need to use the potential difference anywhere right?
No, no V. Capacitance is not dependent on potential.
Isnt capacitance defined as how much charge per unit potential?
Yes, your approach is on the right track, but you do need to consider the potential difference in your calculations. Let's break down the steps to solve this problem.
Step 1: Calculate the area of the inner surface of the erythrocyte.
The formula for the capacitance of a parallel-plate capacitor with a dielectric material in between is given by:
C = (K * ε₀ * A) / b,
where C is the capacitance, K is the dielectric constant, ε₀ is the vacuum permittivity, A is the area of the capacitor plates, and b is the thickness of the dielectric material.
In this case, you are given the value of C = 0.15 pF, K = 5.0, and b = 90 nm (convert to meters, 1 nm = 10^-9 m).
The vacuum permittivity ε₀ is a fundamental constant with a value of approximately 8.85 x 10^-12 F/m.
Using this equation, rearrange it to solve for A:
A = (C * b) / (K * ε₀).
Step 2: Calculate the radius of the erythrocyte.
You can use the formula for the surface area of a sphere A = 4πr^2, where r is the radius of the erythrocyte.
Rearrange the formula to solve for r:
r = sqrt(A / (4π)).
Now, substitute the value of A you obtained in Step 1 into this formula to find the radius r.
Step 3: Calculate the potential difference across the membrane.
Given that the potential difference (V) across the membrane is 94 mV (convert to volts, 1 mV = 10^-3 V).
Step 4: Final calculation.
You now have the radius of the erythrocyte (r) and the potential difference (V). The electric field between the inner and outer surfaces of the erythrocyte is given by E = V / b, where b is the thickness of the membrane.
Using the properties of a spherical capacitor, you know that the electric field inside a capacitor due to a surface change is E = σ / (ε₀ * K), where σ is the surface charge density.
Since the inner surface of the erythrocyte has a positive charge and the outer surface has a negative charge, the surface charge density σ = Q / A, where Q is the charge stored on the capacitor plates (Q = C * V).
The electric field inside the erythrocyte due to surface charge is equal to the electric field between the inner and outer surfaces, so you can set up the equation:
V / b = Q / (A * ε₀ * K).
Now, rearrange this equation to solve for Q:
Q = (V * A * ε₀ * K) / b.
Once you have Q, you can calculate the charge per unit area on the inner surface of the erythrocyte using Q / A.
This process should give you the required information to calculate the radius of the erythrocyte.