# Stuck again - electric field and circular arc

1. May 25, 2004

### PinkFlamingo

Hi there,

I was hoping that somebody could help me. I'm stuck again!

a rod with (lambda) coulombs of charge per meter of its length has the shape of a circular arc of radius R. The rod subtends an angle (theta). Shpw that the magnitude of the electric field at the centre of the circular arc is:

E=[1/(4*pi*E0)] [(2*lambda)/R] sin (theta/2)

Where R is the radius from the rod to the centre of the electic field

OK, I know that I somehow have to change the formula for the electric field of a rod to get that. The formula for the electric field of a rod is:

E==[1/(4*pi*E0)] [(2*lambda)/r]

where r is the vertical distance from the rod

2. May 25, 2004

### turin

Do you know how to integrate?

The electric field for the rod that you gave is for an infinitely long rod. You should first consider whince the formula comes. It is based on Coulomb's Law, superposition, and the definition of the electric field.

To adjust Coulomb's Law, replace the source charge q with dq. This indicates that the charge in Coulomb's Law is just a little piece of some object.

The superposition is the (relatively) complicated issue; it involves integration as well as a cleaver substitution for dq. In general, for a linear charge distribution (a string of charge), you want something of the form: dq = &lambda;(s)ds (where s is the arclength position along the string). I will give you the example for a line of charge along the x-direction: dq = &lambda;(x)dx. Try to figure what it would be for a circular arc. Then, you stick Coulomb's Law with this modification into an integral (over x, for instance). Be careful: the ds may not be the only thing that integrates. You can pull some factors out of the integral, but some must remain as part of the integrand because they depend on the variable of integration.

Then, just divide by the test charge to get the electric field. This can be done in the beginning.