Electric field along a finite rod

[mintsnapple]

Homework Statement

Homework Equations

V = q/(4*pi*E_0*r), when 0 is taken at infinity
dV = -E*ds

The Attempt at a Solution

a. The total charge of the rod is given by Q = lambda*L
So the potential at P is given by
V = (lambda*L)/(4*pi*E_0*y)

b. We can calculate the electric field by differentiating part a with respect to y to get:
E = -(lambda*L)/(4*pi*E_0*y^2)*y-hat

c. Because point P is along the axis of the rod, there is no x-component to the electric field.

 

[Simon Bridge]

Did you have a specific concern?
... for part (a) your answer looks to be the same as if the entire charge were concentrated at the upper end of the rod. Does that make sense?
Try redoing it using calculus and compare.

 

[mintsnapple]

Ah. So I would need to find a dQ value over 1/(4*pi*E_0*y) to integrate?
So, since Q = lambda*L,
dQ = lambda*dL?

And then the integral would go from y to y+L?

 

[Simon Bridge]

You need to be careful about your notation:

Ah. So I would need to find a dQ value over 1/(4*pi*E_0*y) to integrate?

This sentence does not make sense ...
You have to add up the contributions due to each element of charge in the rod.

So, since Q = lambda*L,

dQ = lambda*dL?

If the rod is on a z axis, then the charge between z and z+dz, on the rod, is dQ

And then the integral would go from y to y+L?

No - you only add up the contributions from where there is charge.
If you put the rod in -L<z<0 and point P is at z=y you should find it easier to talk about.

 

[mintsnapple]

Sorry, what I meant was, I need to find V using
dV = dQ/(4*pi*E_0*r). In this case, r is y, and dQ is lambda*dL, that is, dQ is the charge of a really small length along the rod dL.

So then, can I set the bottom of the rod to be 0, and integrate from 0 to L? Why can't I integrate from y to y+L, if I set the top of the rod to be y, so then the bottom must be y+L?

 

[Simon Bridge]

Sorry, what I meant was, I need to find V using
dV = dQ/(4*pi*E_0*r). In this case, r is y,

... no it isn't. r is the distance between point P and the position of the charge element dQ.
"y" is the distance between the end of the rod and the point of interest where you have to find the potential. This distance will be different for different positions along the rod.

So then, can I set the bottom of the rod to be 0, and integrate from 0 to L? Why can't I integrate from y to y+L, if I set the top of the rod to be y, so then the bottom must be y+L?

Oh if you put point P at the origin, then you integrate from y to y+L yes.
The main thing is to be clear about your definitions.

 

[mintsnapple]

Thank you so much for helping me. I am understanding this concept a lot more now.

I just have a few more things:

I get the difference between r and y now. So then, if I say z is the distance between point P and some element dQ, then I also define
dQ = lambda*dz

dV = dQ(4*pi*E_0*r) = (lambda*dz)/(4*pi*E_0*z), integrating from y to y+L, does that sound correct?

 

[Simon Bridge]

dV = [strike]dQ(4*pi*E_0*r)[/strike] = (lambda*dz)/(4*pi*E_0*z), integrating from y to y+L, does that sound correct?

... leave off the middle bit because "r" is no longer in the definition of the problem.
You also want to specify that P is at the origin and you are putting the rod: y < z < y+L.
Now you are set up - need only to do the integration.

 

[mintsnapple]

Thanks! But are part b and c correct? That is for part b, I would just need to differentiate part a with respect to y, and then part c the electric field perpendicular should be 0.

 

[Simon Bridge]

You want to be a tad careful for the last two to use the definition directly.
Particularly for the last one - the answers are not wrong, just not as correct as they could be.

Remember - electric field is a vector, while part (a) gives you a scalar.
Don't forget the reality check - what does the minus sign mean?
Which way should the force point for a positive charge on the y axis?
Is that consistent with your result?

You answer for (c) is a good finesse on the math result - you should still show the same using maths.

Mind you: I've not seen the marking schedule.