Someone had attempted to help me answer this one before, but we sort of hit a de

Question

Someone had attempted to help me answer this one before, but we sort of hit a dead end. Here's the question: A typical cell has a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outside surface, thus making the cell wall a capacitor. What is the capacitance of a 50-um-diameter cell with a 7-nm-thick cell wall whose dielectric constant is 9.0? Because the cell's diameter is much larger than the wall thickness, it is reasonable to ignore the curvature of the cell and think of it as a parallel-plate capacitor. I have asked classmates and apparently you use the surface area of the sphere even though it says to ignore it. I've reached several promising conclusions, but they have not been accepted by mastering physics. Any help would me greatly appreciated! :)

Explanation / Answer

r=d/2 =25um

A=4pir^2 =4pi*(25*10^-6)^2 =7.85*10^-9 m^2

C=KeoA/d =9*(8.85*10^-12)*(7.85*10^-9)/(7*10^-9)

C=8.94*10^-11 F or 89.4pF

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