Seting up the Integral: Electric Potental Energy

[cogman]

I hope that this belongs here, so here is what I would love help with. We are doing Electric potential energy in class right now (and we did electric field prior) but I am struggling with the form the integral needs to take. Once I get it, I can do the math, but it is just understanding all the parts that confuse me.

So here is what I know,

$$V = kq/r$$ is the formula for 2 points

$$V = \int_{surface} k dq/dr$$ I believe (not so sure)

$$dq = \int_{surface} dP/V$$ Where V is volume, area, length.

Is this all correct? Or what am I missing? I would love some examples of how you guys would do it. For my class we are using mostly highly symmetric situations (IE how does a disk, washer, solid sphere effect electric potential of some point perpendicular to) so feel free to make your own up to illistrate the point.

Again thanks for the help. I was unsure on putting this in the homework section as it is not a homework problem I have a question about, but the concept instead.

 

[Claude Bile]

The formula for Electrostatic PE is simple if you have point charges. When you have a charge distribution things get a bit more complicated. Essentially what you want to do is divide a charge distribution into small bits, that way you can apply the original formula to each small bit of charge and add them all up to get the end result.

Mathematically, to divide up a charge distribution use one of the following formula as appropriate. Note that the charge densities, rho, sigma and lambda might vary with position.

Volume: $Q = \int_V \rho d\tau$
Surface: $Q = \int_A \sigma dA$
Line: $Q = \int \lambda dx$

So now (taking volume as an example) you can write;

$$V = k\int_V \frac{\rho}{r}d\tau$$

At this point you have divided up the charge distribution into many small point charges, each with charge $\rho.d\tau$, calculated the contribution to V from each one, then summed them all via the integral.

From here, the problem will progress depending on the symmetry present.

Claude.