Question

The membrane that surrounds a certain type of living cell has a surface area of 4.95×10⁻⁹ m² and a thickness of 9.30×10⁻⁹ m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 5.10. The potential on the outer surface of the membrane is +54.5 mV greater than that on the inside surface. How much charge resides on the outer surface?

Answer 1

To calculate the charge residing on the outer surface of the membrane, we can use the formula for capacitance and the formula for potential difference across a capacitor. The charge is found to be 2.57 x 10^-15 C.

Step-by-step explanation:

The charge residing on the outer surface of the membrane can be calculated using the formula for capacitance: C = ε₀(A/d),

Where C is the capacitance, ε₀ is the permittivity of free space (8.854 x 10^-12 F/m), A is the surface area, and d is the thickness of the membrane.

First, we need to convert the surface area and thickness into meters:

Surface area = 4.95 x 10^-9 m²

Thickness = 9.30 x 10^-9 m

Plugging these values into the capacitance equation:

C = (8.854 x 10^-12 F/m)(4.95 x 10^-9 m²)/(9.30 x 10^-9 m)

Simplifying the equation gives us a capacitance value of C = 4.71 x 10^-12 F.

Next, we can use the formula for potential difference across a capacitor: V = Q/C,

Where V is the potential difference, Q is the charge, and C is the capacitance.

Given that the potential difference is +54.5 mV (converted to volts) and the capacitance is 4.71 x 10^-12 F, we can rearrange the equation to solve for the charge:

Q = (54.5 x 10^-3 V)(4.71 x 10^-12 F)

Calculating the product gives us a charge of Q = 2.57 x 10^-15 C.

Therefore, the charge residing on the outer surface of the membrane is 2.57 x 10^-15 C.