# Potential of a finite line of charge

1. Jun 6, 2013

### Mathoholic!

1. The problem statement, all variables and given/known data

A finite line of charge (L=L1+L2) with a linear density of d(x)=k.x, in which k>0.
This finite line goes from -L2 to +L1 in the x axis.
Calculate the electric field and the electric potential in the point P=(0,H).

2. Relevant equations

dV=(1/(4*pi*ε0))*dq/r

3. The attempt at a solution

I'm trying to solve this problem through the use of the equation of the electric potential above, and calculating this potential for a general point (x,y).

My trouble with this is to determine what's r (the distance from dq to that (x,y) point).

The rest should be easy to deal with once I get r.

Could somebody help with this, please?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 6, 2013

### rude man

r is the distance from your point of observation (0,H) to each differential segment dx of the line.

You know the formula for distance between (x1,y1) and (x2, y2)?

3. Jun 6, 2013

### Mathoholic!

Yes, I know that.

My problem is knowing which differential segment dx to choose on the line (if this makes any sense).

And I'd like to solve this problem from the point of observation of any point (x,y) in the plane.

:/

4. Jun 6, 2013

### haruspex

It will help if you avoid overloading x. If you want to use (x, y) for your general point, take some other variable, u (and correspondingly du) for position along the line of charge. So what's the distance from (u, 0) to (x, y)?

5. Jun 6, 2013

### barryj

haruspex, I am confused by the above statement. Seems to me that if the point is at (0,H) then the radius would be sqrt(x^2 + H^2) but if you used a general point say (X,Y) you would use sqrt((x+X)^2 + Y^2)yes/no?

6. Jun 6, 2013

### haruspex

Mathoholic! has indicated a desire to solve it, not just for the given point (0, H), but for a general position "(x, y)". That's ok (and in fact is no harder as I'll explain), but it will confuse things if x is used as a coordinate there as well as dx being used for an element of the line of charge. I don't care which variable is renamed, but one of them needs to be.
Given that the line of charge is asymmetric about the origin along the x axis (-L2 to +L1), there's nothing special about the x coordinate of P being 0. If you solve it for the given P=(0, H), it's a simple matter to generalise the answer.

7. Jun 7, 2013

### barryj

haruspex, is what I said above true, granted that perhaps we should use different variables (constants) to represent the point, like (U,V)? What do you mean by overloading, I have not heard this term used except in software design.

8. Jun 7, 2013

### barryj

haruspex, I have another question. The problem also asks for the electrical field at a point and I am not totally sure how to do this but I have an idea. Would you break up the E field from the element, dx, into its x and y components, integrate them separately, and then combine the integrated x component and the integrated y component vectorically?

9. Jun 7, 2013

### haruspex

Yes.
I was using it with the same meaning - using one symbol to mean two different things. Sometimes called a pun. Puns can be useful. "y = y(x)" is a pun. The first y denotes a variable, while the second denotes a function.
That would work. Another way is to compute the potential then differentiate.

10. Jun 7, 2013

### barryj

If you computed the potential, then V =f(x,y) so you would have to take partial deraivitaves with respect to x and y then do the vector sum. I think this is the gradient, yes/no?

11. Jun 7, 2013

### haruspex

Yes.

12. Jun 8, 2013

### rude man

OP, have you got V(x,y) yet? I made the point of observation (a,b) since it's a constant in the integration to find V. I integrated wrt x. Then I took the definite integral (limits -L2 and +L1), getting rid of x, and substituted a = x and b = y.

Haven't done E = - grad V yet.