Potential difference and current

[map7s]

Homework Statement

A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 10^7 m. If the potential difference between the inner and outer surfaces of a cell membrane is 75 mV, how much current flows through a square area of membrane 1.3 µm on a side?

Homework Equations

I was thinking to use R=p(L/A) and then V=IR to solve for the current, but I wasn't sure what to use for numbers. I know that V=75 mV and p=1.3 x 10^7, but would L=1.3 micrometers and A=L x 8.0 nm ?

 

[ranger]

First find the resistance, R, by using the formula you wrote down: R=p(L/A). You know the length, L, which is like the thickness. The area, A, is also given and you know the resistivity of the specimen. Whats the problem?

 

[m2003]

Your approach is correct. To solve for the current, you can use Ohm's law, V=IR, where V is the potential difference, I is the current, and R is the resistance. In this case, the resistance is given by the equation R=p(L/A), where p is the resistivity, L is the thickness of the membrane, and A is the area of the membrane.

Therefore, the current flowing through the membrane can be calculated as I = V/R = (75 mV)/(1.3 x 10^7 m)(1.3 x 10^-6 m)/(8.0 x 10^-9 m^2) = 0.0077 mA.

This means that for a square area of membrane with sides of 1.3 micrometers, a current of 0.0077 mA would flow through it when there is a potential difference of 75 mV between the inner and outer surfaces of the membrane.