1. We will model each cell as a parallel platecapacitor with each plate 0.1 mm o
ID: 1747944 • Letter: 1
Question
1. We will model each cell as a parallel platecapacitor with each plate 0.1 mm on a side and the platesseparated by 0.1 mm. The dielectric constant for the cellmembrane is 9.0. Calculate the magnitude of the charge on eachplate to achieve a voltage of 0.15 V. Calculate the magnitudeof the electric field in the cell.
2. Assume that the charge calculated in part 2 isdischarged during a time of 1 ms. Calculate the current(this will actually be the average current). Determine howmany “strings” of capacitors would be needed to givethe typical eel current of 1 A.
Explanation / Answer
side of plate a = 0.1 mm area of the plate A = a ^ 2 = 0.01 mm ^ 2 = 1 * 10 ^ -8 m ^ 2 separation of the plates d = 0.1 mm = 1 * 10 ^ -4m dielectric constant K = 9 Capacitance of the capacitor C = K A / d where = permitivity of free space = 8.85 * 10 ^-12 C ^ 2/ N m ^ 2 substitue values weget C = 79.65 * 10 ^ -4 F voltage V = 0.15 V So, charge on each plate Q = CV = 11.9475 * 10 ^ -4 C the magnitude of the electric field in the cell E = V /d = 15 * 10 ^ 2 V / mRelated Questions
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