# Surface charge density

1. May 22, 2006

### Taryn

I still dont understand from the msg given... another explanation any1

Hey I have two questions... the first one I did I got wrong but am really unsure why... I used sigma=Q/A but didnt get the right answer...
I got that the answer was 1.3277E7 but I am so far off it aint funny!
Please could someone help!
The answer is meant to be 117.5! I dunno tho!

Consider a cell whose cell membrane is 4.7 nm thick and whose resting potential is 62.4 mV. What is the approximate surface charge density (in µ C/m2) on the outside wall of the cell?

Also this question also has me stumped!
Consider the helium atom which consists of two electrons orbiting a nucleus made up of two protons and two neutrons. If both electrons are at a distance of r = 4.5 × 10-11 m away from the nucleus, as in figure, what is the potential energy (in eV) of the helium atom? Treat the nucleus as pointlike. - Give answer to 4 significant figures.

I thought you'd just find the energies of the electron on the helium atom which I got was 1.024E-17 and add them but obviously you dont so if anyone knows wat to do I would be much appreciative, I really just needa know how to do it... or even just a general start!

Last edited: May 22, 2006
2. May 22, 2006

### Taryn

um yeah its cool now... for the first question... I just forgot to times it by 8.85E-12... so if anyone knows wat I am doing wrong for the second one... the help would be great

3. May 22, 2006

### Andrew Mason

Start with a nucleus of +2e and no electrons. Move an electron from infinity to r = 4.5x10-11 m away from the nucleus. Now you have an atom of +e. Move another electron from infinity to r. The work done in moving both electrons is the potential energy of the two electrons.

AM

4. May 22, 2006

### Taryn

I dont understand how that really helps... I know that, but still dont understand where I am goin wrong... could someone help me understand... this is wat I did!
P.E=kqq'/r and then added up wat I got... except I just cant get the right answer???
so does this mean I use the force of a charge and times it by the distance.. your explanation I mean.... but that still doesnt work... The answer is meant to be 112 but I cant get that no matter how hard I try

5. May 23, 2006

### Andrew Mason

I get:

$$PE_1/e = k2e/r = 8.989e9*2*(1.602e-19)/(4.500e-11) = 64.00 eV$$

$$PE_2/e = ke/r = 8.989e9*(1.602e-19)/(4.500e-11) = 32.00 eV$$

So the total potential energy is 96.00 eV.

This is a simplified view of it since it assumes that the electron charge is concentrated at the centre. But it should be very close.

AM