A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius of 2.3 m. What is the magnitude of the electric field at the center of curvature of the arc?

E = 31.329 N/C.

Explanation:

The differential electric field at the center of curvature of the arc is

(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )

where is the radius of curvature.

Now

,

where is the charge per unit length, and it has the value

Thus, the electric field at the center of the curvature of the arc is:

Now, we find and . To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

and this is  

radians.

Therefore,

evaluating the integral, and putting in the numerical values  we get:

higher potential

explanation:

because it is in a high speed so it must move into a of high potential