# Force and Electric field within a cubic lattice

1. Jul 5, 2012

### VictorWutang

1. The problem statement, all variables and given/known data

There are Cl+ ions at the corners of a cube of side a = .4nm and a Cs- ion at the center.
a) What is the net force on the Cs- ion?
b) What is the electric field at the center of the cube cause by all of the Cl ions?
c) What is the electric potential at that point?

2. Relevant equations

F = kq1q2 / r2
E = kq / r
U

3. The attempt at a solution

a) Since each force has an equal force from the opposite corner, I think the net force should be 0. (seems too easy for a test problem, so please let me know if i'm wrong.)

B) For the same reasons as above, wouldn't E be 0?

C) really not sure where to go with this, guessing its an integral relying on the change in distance from each corner going from start position to infinity, and the the sum of those forces for each 8 corners.

Since it could move in a straight line through one of the faces, there would be 2 sets of 4 equal integrals, so i've gotten 4√(3)a/2Edr + 4-√(3)a/2Edr but even if this is right I have E = 0 so I can't do this.

2. Jul 5, 2012

### Yukoel

Hello,
Of course the net electric field at the center is zero.There are symmetrically opposing fields from all vertices.The charges are of the same nature and electric field is a vector.So I think your reasoning for questions a and b is correct. Potential however is a scalar so it has no direction.Just add the potentials due to individual charges ,or for the sake of symmetry seen here multiply potential due to one charge by 8.As far as your integral is concerned (which I will consider a difficult task to work with) see that E=0 only at the center of the cube.(or cuboid)It is not zero anywhere else inside the cube(or cuboid).
(Okay just a side question not related to the topic .Chemistry is not one of my strengths but I think the question should assign negative charges to chlorine ions,isnt it? )
regards
Yukoel

3. Jul 6, 2012

### VictorWutang

so inside the integral is Edr = kq1q2/ r2dr

which, when integrated, is -kq1q2/r

And since ΔV = - ∫Edr so the value i get turns positive, which seems like it would be correct. Then i would just plug in the numbers.

is this right?

4. Jul 6, 2012

### Yukoel

V is a function of the source charge only. What you are deriving is potential energy.Sign seems fine.
You can directly use the expression for potentials considering the point charges separately.Evaluating the integral is correct but unnecessary in your problem.
regards
Yukoel