- #1

enkerecz

- 11

- 0

## 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 [tex]\lambda[/tex] 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 [tex]\theta[/tex] and d[tex]\theta[/tex]

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[tex]\theta[/tex][tex]\lambda[/tex]=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.