# Charge density and electric field

1. Jan 14, 2016

### JFS321

All,

I am wanting to see why charge density divided by e nought is equal to F/q and V/d. Unit cancellation makes it easy to equate F/q = V/d, but why is charge density alone enough to be equal to the electric field? I feel like something is missing here but I can't reconcile it nicely in my head or on paper. To be more clear, I don't understand how we get units of N/C from charge density which is C / m2. Hopefully this makes sense and someone can point me in the right direction.

2. Jan 14, 2016

### DuckAmuck

Charge density is not enough to have an electric field. You also need the distance from your charge.
For point charges, electric field is $$E = k \frac{Q} {d^2}$$
where k is Coulomb's constant, Q is charge, and d is distance.

If you have another charge q at distance d, the force between charges will be F = q*E, that's where E = F/q comes from.
Unit cancellation isn't enough to relate quantities though. V/d only works for point charges. To actually relate V and E, in general, you need to use calculus.

N/C is not a unit for charge density, that is a unit for an electric field. Either way, the units don't make sense to you because you are forgetting Coulomb's constant (the k in the formula above), which has units of Nm^2/(C^2).
Depending on what unit system you use, you can express electric field with or without this constant, but it changes you units for charge.

Last edited: Jan 14, 2016
3. Jan 14, 2016

### JFS321

Ok, I follow the units here. What am I missing to show that sigma / e nought is equivalent to F / q...?

4. Jan 14, 2016

### nasu

$\epsilon_0$has his own units. Did you consider this?
It's not charge density alone.