The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 5.41 10-9 m2, a plate separation of 8.5 10-9 m, and a dielectric with a dielectric constant of 4.5.

What is the energy stored in such a cell membrane if the potential difference across it is 0.0679 V?

? J

Energy stored in cell membrane is 5.83 x 10⁻¹⁴ J.

Explanation:

Given :

Area of parallel plate capacitor,A = 5.41 x 10⁻⁹ m²

Distance between the plates,d = 8.5 x 10⁻⁹ m

Dielectric constant, k = 4.5

Capacitance of a parallel plate capacitor is given by the relation :

C = (kε₀A)/d

Here ε₀ is vacuum permittivity and its value is 8.85 x 10⁻¹² F/m.

Substitute the suitable values in the above equation.

C = 2.53 x 10⁻¹¹ F

Potential difference between the plate, V = 0.0679 V

Energy stored in a capacitor is given by the relation :

Substitute the values of C and V in the above equation.

E = 5.83 x 10⁻¹⁴ J

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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when it gains or loses electrons