# Point charges and electric potential

1. Sep 25, 2010

### fawk3s

I have some difficulties grasping the idea of this equation

Wp = q * E * r

Where
Wp - electrical potential energy
q - charge
E - the strength of the electric field
r - radius, or distance from the other point charge

So basically, say we have 2 point charges. One positive, and one negative, so they attract.
But I cant imagine how the work done by the moving point charge is the same we get by calculating by that formula. Because the closer the point charge gets to the other point charge, the bigger the E gets. So the electric force applied on that point charge isnt constant. Its getting bigger.
My physics textbook also says that the way to calculate the work done by the point charge in this situation is too difficult to describe in a highschool textbook. But yet they still give me that equation. Why?

I can imagine the same thing with Earth and its gravitational potential energy, but the change in g (9,81 m/s2) is so small over short distances that there's no point to really observe it.

Or am I getting this whole thing wrong? Please help !

Last edited: Sep 25, 2010
2. Sep 25, 2010

### Idoubt

Im going to assume you know calculus ( if not, integration is the continuous summing of a variable)

The potential energy is the same as the work done to bring the particle to that point from an infinte distance ( ie a distance where the potential energy is 0 )

The electric field E is the force acting on 1 C of charge (sign sensitive ) so q*E gives the force acting on the charged particle.

The work done by this force is given by the intergral of F.dr ( the component of the force in the direction of displacement times the displacement )

Now the equation q*E*r is the case in which F ( ie the electric field E ) is constant over the distance r ( which can be approximately true in some cases though not for point charges )

then the integral becomes F*r = q*E*r

To obtain the work done for 2 point charges we have to integrate E and r which will get complicated as you apply the by-parts integration ( both E and r are functions of r )

hope this helps

3. Sep 25, 2010

### fawk3s

Yes, I understand how the equation works in a homogeneous electric field, and they actually gave us Wp = q * E * d for a homogeneous electric field in the textbook, but Wp = q * E * r for point charges. I dont quite understand what to do with the latter since I have no idea how to use it with a changing E. They are basically the same though.

4. Sep 25, 2010

### granpa

thats what calculus is for.
if you want to do physics then you MUST learn calculus.

conceptually you break the changing field into infinitesimal unchanging parts.
Calculate the energy in each part.
Then add them all up (integrate them)

the integral of an inverse square law will give you an inverse first power law
therefore the energy of an infinitesimal charge falling from infinity
into a finite (unmoving) charge will be proportional to 1/d
where d is the distance from the finite charge.

Last edited: Sep 25, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook