TP 21.14

Given: l = 8.00 nm

r = 1.3 x 10⁷ W-m

a = 1.00 mm

DV = 75.0 mV

(a) Find: I - ?

Ohm's Law:

DV = I R

R = r l /A = (1.3 x 10⁷ W-m)(8.00 x 10^-9 m)/(1.00 x 10^-6 m)² = 10.4 x 10¹⁰ W

I = DV/R = (75.0 x 10^-3 V)/(10.4 x 10¹⁰ W) = 7.21 x 10^-13 A

(b) If the thickness of the membrane is doubled, the resistance of the membrane goes up by a factor of two. If the potential difference remains the same, the current drops by a factor of two.