Integrating a charge along a line

1. Feb 9, 2010

burns12

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

total charge Q = 200[nC] is spread along a line from x=−100[mm] to x=+100[mm] .
a) write the dV caused by each dQ=λ dx .
b) integrate along the line of charge ; write V as a function of y .
c) take the derivative with y , to obtain that component of the Electric (vector) field .
d) take the derivative with y , again , to find where (along y-axis) the Ey is maximum

2. Relevant equations
We've used potential energy U=1/2QV...V=kQ/r...that's about it.

3. The attempt at a solution
What I can't get started with is the writing dV caused by each dQ. I'm not even sure what that means. And how can I write V as a function of y and take its derivative if there are no y's in any of these equations?

Any guidance here would be awesome, thanks

2. Feb 9, 2010

rl.bhat

hi burns12, welcome to PF.
Take a small element dx at a distance x from the origin.
Consider a point P at a distance y from the origin along y-axis.
Charge dq in the dx element is λ*dx, where λ = Q/L.
As you have mentioned
dV = k*dq/r. Put the value of dq and r.
To find V at P, take the integration between the limits x = -0.1 m to 0.1 m.

3. Feb 9, 2010

burns12

Thanks!

Ok so, the y comes into play as a distance on the y axis that this dq charge is going to affect?
And do I integrate the dV = k*dq/r ? Or do I sub in λdx for dq and then integrate with respect to x to find dV?
I'm still not sure where this distance y would come into this.

4. Feb 9, 2010

rl.bhat

dq = λ*dx and r = sqrt( x^2 + y^2)

5. Feb 9, 2010

burns12

Alright awesome I think I got it. Is the r = to sqrt(x^2 +y^2) because it's like the components of r in a sense?

6. Feb 9, 2010

Yes.

7. Feb 9, 2010

burns12

Awesome, thanks for your help, that had me stumped for 2 days now ha