Originally Posted by **tag16** A charge Q is arranged evenly on a wire bent into an arc or radius R as shown in the Figure. What is the electric field at the center of the arc as a function of the opening angle theta? Sketch a graph of the electric field as a function of theta for 0<theta<180 degrees. http://i848.photobucket.com/albums/a...g?t=1259862072
E= KQ/R^2 S cos theta (the S is suppose to be an integral symbol)
Is the above equation correct? If not, I'm not really sure what to do, also I'm not sure how the graph should look. |

no dear.. if you consider this wire to be a straight wire then the electric field due to this wire must be

E=Q/4*pi*e*x^2*(x^2+a^2)^1/2

where

E=electric field,Q=charge,e=8.85x10^-12,x=length of wire,a=distance of wire and a test charge..

where in your case a=R(radius of the semicircle)..

just try over it