How Do You Calculate the Electrical Properties of a Cell Membrane?

[ussrasu]

The membrane of a cell is electrically equivalent to a parallel plate capacitor. A typical cell has a spherical shape with a radius 10 μm and it has a potential of −60 mV with respect to outside. The thickness of the membrane is about 0.01 μm, and it has a dielectric constant of 2.
Find:
(a) the capacitance,
(b) the magnitude of the electric field in the membrane,
(c) the stored electrical energy, and
(d) the magnitude of the separated charge.

Where to start on this question?

 

[Nate810]

a) The capacitance of a parallel plate capacitor is given by C = εA/d, where ε is the dielectric constant, A is the area of the plates, and d is the distance between the plates. For a spherical cell, the area of the plates is 4πr^2, so the capacitance is:C = (2 * 4πr^2)/(0.01 μm) = 8πr^2/0.01 μmb) The electric field in the membrane can be calculated using the equation E = V/d, where V is the potential difference across the membrane and d is the thickness of the membrane. Thus, the electric field in the membrane is:E = −60 mV/ 0.01 μm = -6000 V/mc) The stored electrical energy is given by U = ½CV^2, where C is the capacitance and V is the potential difference across the membrane. Thus, the stored electrical energy is:U = ½(8πr^2/0.01 μm)(-60 mV)^2 = 9.6 x 10^-3 Jd) The magnitude of the separated charge is Q = CV, where C is the capacitance and V is the potential difference across the membrane. Thus, the magnitude of the separated charge is:Q = (8πr^2/0.01 μm)(-60 mV) = 4.8 x 10^-5 C

 

[quicknote]

To answer this question, we need to use the formula for capacitance, which is C = εA/d, where C is capacitance, ε is the dielectric constant, A is the area of the plates, and d is the distance between the plates.

(a) To find the capacitance, we first need to calculate the area of the membrane. Since the membrane is spherical, we can use the formula for the surface area of a sphere, which is A = 4πr^2, where r is the radius of the sphere. Plugging in the given values, we get A = 4π(10μm)^2 = 400π μm^2. Now, we can plug in the values for ε and d to get the capacitance: C = (2)(400π μm^2)/(0.01μm) = 80,000π μF.

(b) To find the magnitude of the electric field in the membrane, we can use the formula E = V/d, where E is the electric field, V is the potential, and d is the distance between the plates. Plugging in the given values, we get E = (-60mV)/(0.01μm) = -6,000 V/μm.

(c) To find the stored electrical energy, we can use the formula U = (1/2)CV^2, where U is the stored energy, C is the capacitance, and V is the potential. Plugging in the values, we get U = (1/2)(80,000π μF)(-60mV)^2 = 2,160π μJ.

(d) To find the magnitude of the separated charge, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential. Plugging in the values, we get Q = (80,000π μF)(-60mV) = -4,800π μC.

In summary, the capacitance of the cell membrane is 80,000π μF, the magnitude of the electric field is -6,000 V/μm, the stored electrical energy is 2,160π μJ, and the magnitude of the separated charge is -4,800π μC.