Electric field due to a Linear charge density

[enkerecz]

Homework Statement

Two long, thin parallel rods, a distance 2b apart, are joined by a semicircular piece of radius b, as shown. Charge of uniform linear density \lambda is deposited along the whole filament. Show that the field E of this charge distribution vanishes at the point C. DO this by comparing the contribution of the element at Ato that of the element at B which is defined by the same values of \theta and d\theta


Note: Point C is the center of the semicircle which is the connector of the two rods. The segments d(theta) cut out pieces along the rods.

Homework Equations

rd\theta\lambda=dQ1

The Attempt at a Solution

Okay, so I do not know how to do this. First, point B lies on the same side of the rod as A, but is along the straight piece, i.e. prior to the semi circle attachment.

The first relevant equation I have given allows you to calculate the charge for Q at A, thus

allowing you to calculate E1. B, on the other hand, because it lies on the straight, is not so

easily calculated. Of course, geometrically, A and B have an angle theta above and below

the diameter of the semi-circle and also share the same angle d(theta) By extending the

sectors given by d(theta) through to the opposite sides, it becomes obvious that there is

symmetry, i.e. equal sectors are cut out with equal magnitudes but opposite signs, and the Electric fields due to these points would cancel out because the Electric fields of both are in opposing directions.

Like I said earlier, I can only solve for the point A because the radius, r, is clearly b. As for Sector B, it makes a right triangle with base b, but neither the longer leg nor the hypotenuse are given.. Thus, I can only explain geometrically, but not mathematically, and I'm not sure how the question is made to be answered.

 

[SammyS]

Homework Statement

Two long, thin parallel rods, a distance 2b apart, are joined by a semicircular piece of radius b, as shown. Charge of uniform linear density \lambda is deposited along the whole filament. Show that the field E of this charge distribution vanishes at the point C. DO this by comparing the contribution of the element at A to that of the element at B which is defined by the same values of \theta and d\theta


Note: Point C is the center of the semicircle which is the connector of the two rods. The segments d(theta) cut out pieces along the rods.

Homework Equations

rd\theta\lambda=dQ1

The Attempt at a Solution

Okay, so I do not know how to do this. First, point B lies on the same side of the rod as A, but is along the straight piece, i.e. prior to the semi circle attachment.

The first relevant equation I have given allows you to calculate the charge for Q at A, thus

allowing you to calculate E1. B, on the other hand, because it lies on the straight, is not so

easily calculated. Of course, geometrically, A and B have an angle theta above and below

the diameter of the semi-circle and also share the same angle d(theta) By extending the

sectors given by d(theta) through to the opposite sides, it becomes obvious that there is

symmetry, i.e. equal sectors are cut out with equal magnitudes but opposite signs, and the Electric fields due to these points would cancel out because the Electric fields of both are in opposing directions.

Like I said earlier, I can only solve for the point A because the radius, r, is clearly b. As for Sector B, it makes a right triangle with base b, but neither the longer leg nor the hypotenuse are given.. Thus, I can only explain geometrically, but not mathematically, and I'm not sure how the question is made to be answered.

Hi enkerecz.

Since I don't have the figure, I'll assume that points A, B, and C are co-linear, and that the line they lie on makes an angle θ relative to the line that passes through point C and is parallel to the line charges.

How far from point C is point B?

How long is the length of rod intercepted by dθ at point B? How much charge is there in this length of rod? What is the magnitude of the electric field, E₂ at C, due to this length of charged rod?

 

[enkerecz]

okay, I am not sure if you have done one of the window optimization problems in calculus, but the shape is similar to that. Two straight rods with a distance 2b between them. The two rods are connect by a semicircular window. B is the length of rod and it is shaped like a parallelogram, as it has been shaped by d(theta) which runs from the mid-point C of the semicircle down the left side. A lies at an arbitrary point along the semi-circle and is a segment that is also shaped by d(theta). If you were to draw a dotted diameter through C, you can see that a right triangle is made near b, which means the base is bcos(theta) and the hypotenuse is the inner part of the d(theta) segment. If drawn correctly, on the left hand side there will be sector A, a radius b and theta goes between the imaginary diameter and that d(theta) for A.

Right below the diameter, you will have the right triangle and theta is there as well. Theta=Theta. d(theta)=d(theta), which is why it is seen easily geometrically.

Q for A is r(d(theta))Lambda, but I think I miscalculated Q for B because I used
Q_B=R(d(theta))Lambda, where big R is bcos(theta)

Sorry for being so verbose, I'm trying to get this through as clearly as possible :(**EDIT** B is not the length of the rod, the rod length is unknown. B is the the designated slice of "length" cut out by the segment d(theta) which originates at C. The shape of B is a parallelogram, which leaves me clueless on estimating anything about it, especially when given a linear density.

Also, the only length given is that the two rods are separated by distance 2b, thus giving the semi circle radius b.

 

[enkerecz]

This is an image of what I'm working with

 

[SammyS]

Here's a Figure which may help.

http://usera.ImageCave.com/DGSamSnyder/linear charge1.jpg