# Understanding some fundamentals - electrostatics

1. Sep 18, 2007

### Z COOL

I was reading my text and doing some problems and I realize that I am having a lot of issues understanding some of the basic stuff :(

I have a few questions about charge, fields etc. Like given the charge on an object after charge has been transferred by rubbing, how do you tell how many electrons have been transferred to the newly charged object? Do I use the equation for charge (q = [Np - Ne]e) ?

I was also reviewing E = F/q and am wondering:
- if E = F/q , can we say that this is because the force gets 'spread out' over the size of the charge? How would I draw how the force is acting upon a point charge?
- If I play around with this equation, I notice that when the charge is small, the electric field is big, and vice versa. Does this imply that a smaller electric field or force will be felt on a larger charge? Can a large point charge feel a large force / field?

Basically, I'm having trouble picturing what's going on in these two situations on the micro-level. (PS I hope this isn't in the wrong topic - I'm new to the forum :) )

2. Sep 18, 2007

### fantispug

You've got the right idea for charge transferred by rubbing - if N electrons are transferred by rubbing, the object transferred from will gain a positive charge q=N*e, and the object transferred to will get an equal negative charge. (Protons don't tend to move around). So the number of electrons is just N=q/e.

The electric field E is a way of measuring the force (per unit charge) of some distribution of charge at some point in space.

Be careful with the equation E=F/q; it is saying:
The electric force E at a point in space, is defined as the force on a (point) charge placed at that point, divided by its charge q (ignoring the effect of this charge - a charge does not classically act on itself).
So since we are talking about a point charge q, here talking about the "size of the charge" does not make good physical sense.
For a real distribution of charge, e.g. a sphere of charge, since the electric field is continuous, the force is spread out in some way over the body. You can draw the force on each point charge in the distribution as an arrow, and the total force on this body is the vector sum of all these arrows, acting through the centre of mass.

(If you use a continuous distribution, it will become the force on every infinitesimal piece of charge dq, and the sum will become an integral)

If the charge is small, the electric field is big, for a given force. This implies that if we want to exert the same force on a smaller charge we need a bigger electric field. (Remember: the charge can't affect the electric field at the point it is located! So E at that point is unaffected by the size of the charge.)
So F=Eq implies that a larger charge will experience a GREATER force for a given electric field. A small point charge can experience a large force but it requires a greater electric field.