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Boozehound
Apr19-07, 11:24 PM
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 dont know if i am using the wrong formulas. most likely thats the case. so if you see something im doing wrong and can point me in the right direction that would be a great help. thanks.

denverdoc
Apr19-07, 11:46 PM
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.