The membrane that surrounds a certain type of living cell has a surface area of 4.50 × 10-9 m2 and a thickness of 1.00 × 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.3. The potential on the outer surface of the membrane is +62.0 mV greater than that on the inside surface. If the charge is due to positive ions (charge +e), how many such ions are present on the outer surface?
q = kappa epsilon A V/d
q = 5.3* 8.8e-12 * 4.5e-9 *62e-3 / 1e-8
Once you have this number the charge of a single ion (1.67e-19) to find out how many charges you need.
To find the number of positive ions present on the outer surface of the membrane, we need to use the formula for capacitance of a parallel plate capacitor. The formula is:
C = (ε₀ * εᵣ * A) / d
Where:
C is the capacitance of the capacitor,
ε₀ is the permittivity of free space (vacuum) (8.85 × 10⁻¹² F/m),
εᵣ is the relative permittivity or dielectric constant of the material,
A is the surface area of the capacitor plates,
d is the distance or thickness between the plates.
Given data:
A = 4.50 × 10⁻⁹ m² (surface area of the membrane),
εᵣ = 5.3 (dielectric constant),
d = 1.00 × 10⁻⁸ m (thickness of the membrane).
First, substitute the given values into the formula:
C = (8.85 × 10⁻¹² F/m * 5.3 * 4.50 × 10⁻⁹ m²) / (1.00 × 10⁻⁸ m)
Simplifying the equation:
C = 2.37 × 10⁻¹² F
Now, we can use the equation for capacitance:
C = Q / V
Where:
C is the capacitance of the capacitor (2.37 × 10⁻¹² F),
Q is the charge on the capacitor plates,
V is the potential difference across the plates (62.0 mV = 62.0 × 10⁻³ V).
Rearranging the equation to solve for Q:
Q = C * V
Substitute the values:
Q = 2.37 × 10⁻¹² F * 62.0 × 10⁻³ V
Calculating the charge (Q):
Q = 1.47 × 10⁻¹⁴ C
Since each positive ion carries a charge +e, where e is the elementary charge (1.6 × 10⁻¹⁹ C), we can calculate the number of positive ions:
Number of positive ions = Q / e
Substitute the values:
Number of positive ions = 1.47 × 10⁻¹⁴ C / 1.6 × 10⁻¹⁹ C
Calculating the number of positive ions:
Number of positive ions ≈ 9.19 × 10⁴
Therefore, there are approximately 9.19 × 10⁴ positive ions present on the outer surface of the membrane.