Cell Membrane Electric Field

[Boozehound]

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.0730 V exists across the membrane. The thickness of the membrane is 8.22E-9 m. What is the magnitude of the electric field in the membrane?

so i took .0730V and multiplied it by 1.60E-19C to get an answer in joules. i ended up getting 1.168E-20J. i then took that and divided it by .0730V to get an answer in coulombs. and i got 1.6E-19C for that. then i took my answers and plugged them into the equation. E=8.99E9(1.6E-19)/(8.22E-9)^2. then the final answer i got was . and i punched it in and it was wrong. i don't know if i am using the wrong formulas. most likely that's the case. so if you see something I am doing wrong and can point me in the right direction that would be a great help. thanks.

 

[denverdoc]

wait, too hard, and I'm thinking youre confused re maybe the notion of a test charge. I think, iirc, you just need to treat this as a ccapacitor which relates E,V, and thickness.

 

[m2003]

Hello! I can provide some clarification on your calculations. First, it is important to note that the electric field is measured in units of volts per meter (V/m). So, to find the magnitude of the electric field in the membrane, we need to divide the potential difference by the thickness of the membrane (in meters). This will give us the electric field in units of V/m.

So, using your values, the electric field in the membrane would be calculated as follows:

Electric field = (0.0730 V) / (8.22E-9 m) = 8.88E6 V/m

This means that the electric field in the membrane has a magnitude of 8.88 million volts per meter. This is a very strong electric field, which is necessary for many cellular processes to occur.

I hope this helps clarify your calculations. Keep up the good work in your studies of science!