Question

The fluids inside and outside a cell are good conductors separated by the cell wall, which is a dielectric. Thus the cell has capacitance; charge may be stored on its inner and outer surfaces (see the figure below). It is a good approximation to treat the thin charged layer as a parallel-plate capacitor. Typically the wall is 9.50 x 10-9 m thick and has a dielectric constant of 5.00. (everybody's responses to this question before were incorrect so I need an explanation as to how you got your answer)

(a) Find the capacitance per cm2 of cell membrane. EXPLAIN PLEASE




(b) Suppose the potential difference across the cell wall is 0.103 V. Find the magnitude of the charge stored on either side of the cell wall per cm2.


(c) Find the field strength across the membrane.

Answer 1

Capacitance is given by, C-504 (5)(8.85x10-12 F/m 9.50x109 m 4.657 x 10-3 F/m2 , 4657 x10% (10-2)F/cm 2 = 4.657×10-7 F/cm 2 Therefore, the capacitance is 4.657x10 F/cm2 Magnitude of charge is given by, -CV (0.103 V)(4.657x10 4.79×10-8 C/cm2 Flem2) Magnitude of charge is 4.79x10-8 C/cm2 Electric field strength is given by, 0.103 V 9.50x10-9rm -10.84x106 V/m Therefore, the electric field strength is 10.84x106 V/m