E Field Calc: 2 Charged Line Segments

[Corneo]

2 charged line segments are placed on the y axis. L which is between $a \leq y \leq b$, and L' $-b \leq y \leq -a$. Find the electric field at the point (0,0,h). L has line charge density $\rho$ and L' has $- \rho$.

I believe that the z component of of the electric field will be canceled out by symmetry. The only component is the y component. In this particular example, the component will be in the negative y direction that is $\vec E = - E_y \vec j$

I plan to find the electric field generated by just one of the line segments and multiplying by 2.

$$\displaystyle{ <br /> \vec E = 2 \frac {1}{4 \pi \epsilon} \int_{-b}^{-a} \frac { - \rho y \vec j}{(h^2+y^2)^{\frac {3}{2}}}} = \frac {- \rho}{2 \pi \epsilon} \bigg( \frac {1}{\sqrt{b^2+h^2}} - \frac {1}{\sqrt{a^2+b^2}}\bigg) \vec j$$.

The thing that troubles me is that $$\frac {1}{\sqrt{b^2+h^2}} - \frac {1}{\sqrt{a^2+h^2}} < 0$$. E actually points in the positive y direction. What is wrong here?

 

[jollygood]

this is a 3D problem. since your wires have finite lenghts, you need to have all three coordinates (x,y,z dependence) in your general solution.
Once you have the general solution, then apply necessary conditions to get the field at the point you need.

it will be easier if u use cylindrical coordinates. take wires to be along z direction.

 

[Corneo]

this is a 3D problem. since your wires have finite lenghts, you need to have all three coordinates (x,y,z dependence) in your general solution.
Once you have the general solution, then apply necessary conditions to get the field at the point you need.

it will be easier if u use cylindrical coordinates. take wires to be along z direction.

Can you elaborate more on what you just said?

I can't change the coordinate system because later on in the problem I am introducing a ring of radius R. The ring sits on the xy plane with the center at the origin, and I want to find the electric field generated by this ring also.

 

[xman]

$$\displaystyle{ <br /> \vec E = 2 \frac {1}{4 \pi \epsilon} \int_{-b}^{-a} \frac { - \rho y \vec j}{(h^2+y^2)^{\frac {3}{2}}}} = \frac {- \rho}{2 \pi \epsilon} \bigg( \frac {1}{\sqrt{b^2+h^2}} - \frac {1}{\sqrt{a^2+b^2}}\bigg) \vec j$$.

The thing that troubles me is that $$\frac {1}{\sqrt{b^2+h^2}} - \frac {1}{\sqrt{a^2+h^2}} < 0$$. E actually points in the positive y direction. What is wrong here?

Follow the above advice, but also note from your calculation, since $$-b<-a \Rightarrow \frac {1}{\sqrt{b^2+h^2}} - \frac{1}{\sqrt{a^2+h^2}} > 0$$. Since, b is a smaller number then one over a small number is less than one over a large number. Imagine, b=-1, a=-2, h=whatever

$$\frac{1}{\sqrt{(-1)^{2}+h^{2}}>\frac{1}{\sqrt{(-2)^{2}+h^{2}}$$

So what you have in parenthesis is actually greater than zero and since there's a minus this is actually in the $$-\hat{y}$$ direction. Also a hint on the problem it may help to consider the electric field lines, and drawing a picture, always helps me anyway. Hope this helps sincerely, x

 

[Corneo]

But b is greater that a in the problem. The two finite wires are located at

$$a \leq y \leq b$$ and $$-b \leq y \leq -a$$

BTW I meant to say the ring will sit on the xy plane.

 

[xman]

But b is greater that a in the problem. The two finite wires are located at

$$a \leq y \leq b$$ and $$-b \leq y \leq -a$$

BTW I meant to say the ring will sit on the xy plane.

Right if you are looking at $$y>0$$ for which case the term in parenthesis is negative and a negative times a negative is a positive, i.e. in this case points in the $$\hat{y}$$ direction. In other words, your calculation is correct, you're just misinterpreting the signs.

 

[Corneo]

But I was integrating over the line segment which has the negative rho (Assuming rho is positive). The E field from this segment should be in the $- \hat y$ direction.

All that aside. If I now introduce a circular ring with radius a centered on the origin and sitting on the xy axis. The E field from the ring is strictly in the $\hat z$ direction (assuming it's rho > 0). So there is no way I can chose the line charge density of the wires in such a way to cancel the E field from the ring right?

 

[xman]

But I was integrating over the line segment which has the negative rho (Assuming rho is positive). The E field from this segment should be in the $- \hat y$ direction.

All that aside. If I now introduce a circular ring with radius a centered on the origin and sitting on the xy axis. The E field from the ring is strictly in the $\hat z$ direction (assuming it's rho > 0). So there is no way I can chose the line charge density of the wires in such a way to cancel the E field from the ring right?

Yeah, I'm following what you're saying, and for the first part of the problem we should expect the E field to be in the minus y direction right, since the minus line charge field lines draw into the charge, and the positive also point this direction. As per your second question, are the line charges variable? and you have to determine if there is a way to make up, with the geometry stated if there is a way to make the E field at the given point zero? is that it? if so, i would imagine if both line charges are negative, then their horizontal components would cancel, and (assuming the linear charge density of the ring is positive) the two line charges would produce a negative E field at the point P and the ring would generate a net positive pointing up. Hence, the problem then becomes to choose the "strength" of the negative charge such that the E field from the two line of charges exactly cancels the E field of the ring. Does this make sense? In other words, yeah it should be possible. I haven't worked it out, but if you get stuck, I can.

 

[Corneo]

Great, thanks for all the help. I understand now.