Potential difference between two points problem

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wheybags
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Homework Statement



Given a point charge of 500pC at the origin, find the potential difference between points a and b at distances 5m and 15m respectively along the x-axis.

Homework Equations



Coulomb's law, I think?

The Attempt at a Solution



The reason I thought coulombs law was relevant was because it was mentioned in the first part of the equation. I tried to apply Coulomb's law, but got nowhere, as I don't have a charge for a and b, and also, I think that would ignore the charge at the origin, so it was probably the wrong way to go about it. To be honest, it just doesn't seem to me like there is enough information given to do anything with it, really.
All help gratefully received.
 
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Have you learned anything called as electrostatic potential??
If not, read some book and grab an idea of it. It will help you to understand the problem!
 
I looked, but couldn't find anything really. Wouldn't I need to know the charge at a and b to do anything?
 
wheybags said:
I looked, but couldn't find anything really. Wouldn't I need to know the charge at a and b to do anything?

It's a point charge, so no.

Change in electric potential is defined as [tex][V(\mathbf{r_2})-V(\mathbf{r_1})] = -\int_{\mathbf{r_1}}^{\mathbf{r_2}} \!\!\! \mathbf{E} \cdot d\mathbf{r}[/tex]

What is the electric field for a point charge?
 
r1 and r2 are distances from the origin, right? But what is V? also, is E the electric field, which is F/q? If so, I would need q, the charge at each point, wouldn't I? but you said I didn't.
 
I think I worked this out, would appreciate if someone could confirm my answer.

500pC = 500x10^-12 C
Potential at a point = Kₑ (q/r)
applied these and got:
Va = 9x10^-12
Vb = 3x10^-10
Potential difference between a and b = |Va - Vb|
=> 2.91x10^-10
 
Oh god, I'm an idiot. I was using Ke = 9...