A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 × 107Ω.m.
(i) If the potential difference between the inner and outer surfaces of a cell membrane is 75 mV, how much current flows through a square area of membrane on a side?
(ii) Suppose the thickness of the membrane is doubled, but the resistivity and potential difference remain the same. Does the current increase or decrease? By what factor?
decrease by 2
To answer both parts of the question, we can use Ohm's Law, which states that the current flowing through a conductor is directly proportional to the potential difference across it and inversely proportional to its resistance.
(i) First, let's calculate the resistance of the cell membrane. The resistance (R) of a conductor is given by the formula R = resistivity (ρ) x length (L) / cross-sectional area (A).
Given:
Cell membrane thickness (L) = 8.0 nm = 8.0 x 10^(-9) m
Resistivity (ρ) = 1.3 x 10^7 Ω.m
The cross-sectional area (A) of the membrane is not explicitly given, but we can assume it to be a square with one side equal to the thickness of the membrane.
Therefore, A = (8.0 x 10^(-9) m) x (8.0 x 10^(-9) m) = 64 x 10^(-18) m^2
Now, we can calculate the resistance:
R = (1.3 x 10^7 Ω.m) x (8.0 x 10^(-9) m) / (64 x 10^(-18) m^2)
R = 1.6 x 10^9 Ω
Next, let's calculate the current flowing through the square area of the membrane. Using Ohm's Law (I = V / R), where V is the potential difference and R is the resistance:
Given:
Potential difference (V) = 75 mV = 75 x 10^(-3) V
I = (75 x 10^(-3) V) / (1.6 x 10^9 Ω)
I ≈ 4.69 x 10^(-11) A
Therefore, the current flowing through the square area of the membrane is approximately 4.69 x 10^(-11) A.
(ii) If the thickness of the membrane is doubled while keeping the resistivity and potential difference the same, we need to determine how this affects the current.
Let's calculate the new cross-sectional area (A) of the membrane after doubling the thickness:
New thickness (L') = 2 x 8.0 x 10^(-9) m
New A = (2 x 8.0 x 10^(-9) m) x (2 x 8.0 x 10^(-9) m) = 4 x 64 x 10^(-18) m^2
New A = 256 x 10^(-18) m^2
Now, we can calculate the new resistance (R') of the membrane using the same resistivity:
R' = (1.3 x 10^7 Ω.m) x (2 x 8.0 x 10^(-9) m) / (256 x 10^(-18) m^2)
R' = 1.6 x 10^9 Ω
We can see that the resistance remains the same even after doubling the thickness.
Finally, let's calculate the new current (I') flowing through the square area of the doubled-thickness membrane using Ohm's Law:
I' = (75 x 10^(-3) V) / (1.6 x 10^9 Ω)
I' ≈ 4.69 x 10^(-11) A
As we can observe, even after doubling the thickness of the cell membrane, the current remains the same. So, the current does not increase or decrease. It stays constant.
In summary,
(i) The current flowing through a square area of the membrane is approximately 4.69 x 10^(-11) A.
(ii) Doubling the thickness of the membrane does not change the current flowing through it. The current remains constant.