Integrating a charge along a line

[burns12]

Homework Statement

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

Homework Equations

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

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

 

[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.

 

[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.

 

[rl.bhat]

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.

dq = λ*dx and r = sqrt( x^2 + y^2)

 

[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?

 

[rl.bhat]

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?

Yes.

 

[burns12]

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