Uniformly charged rod in an arc

[StephenDoty]

If a rod with charge density lambda is bent into a 3/4 circle what is the E field at the center?


Well if the rod is 3/4 of a circle then neither the x or y components can cancel out, thus

E= k*lambda/ R (i) + k*lambda/R (j)

right?

 

[Defennder]

You need to tell us how the circular arc is oriented. And I don't see how you got that. I got a different answer for magnitude.

 

[StephenDoty]

from pi/2 to 2pi

meaning the test particle would go up and to the right.

as I drew the lines of E I couldn't tell how any component could cancel, and if you spit it up neither the x or y component cancel out as the r's are different. If the rod was a half circle the y's would cancel out but if it is just a 1/4 circle there is an x- and y-components, thus for a 3/4 circle there would be x- y- components.

and if you add another 1/4 of a circle you would get a ring at which there is no E field at the center. and if you draw two dq lines from opposite sides of the 3/4 circle arc the resultant would be going up and to the right.

After that I just integrated x and y components from pi/2 to 2pi as my instructor indicated.

 

[alphysicist]

from pi/2 to 2pi

meaning the test particle would go up and to the right.

as I drew the lines of E I couldn't tell how any component could cancel, and if you spit it up neither the x or y component cancel out as the r's are different. If the rod was a half circle the y's would cancel out but if it is just a 1/4 circle there is an x- and y-components, thus for a 3/4 circle there would be x- y- components.

You can redefine your axes so that one set of components will cancel. For example, if you define the resultant field to be along the y-axis, so that the arc is from (3/4)pi to (1/4)pi, all of the x components will cancel.