# Help with electromagnetism

mateoguapo327
help! I need help with this homework problem.

--A line of positive charge is formed into a semicircle of radius R=60.0 cm. The charge per unit length along the semicircle is described by the expression [lamb]= [lamb]naught cos [the] . The total charge on the semicircle is 12.0 microcoulombs. Calculate the total force on a charge 3.00 microcoulombs at the center of curvature.--

The figure shows the semicircle with is center at the origin going from 0 to pi. [the] is the angle formed by dragging R clockwise from the positive y axis.

consider the electric field at the center of the semicircle due to a little piece of the semicircle at angle &theta;. by symmetry, only the vertical component of this field contributes:

dEy = - 1/4&pi;&epsilon;0 cos &theta; dq/r2 = - 1/4&pi;&epsilon;0 cos &theta; &lambda;rd&theta;/r2 = - 1/4&pi;&epsilon;0 cos2 &theta;&lambda;0d&theta;/r

then integrate from &pi;/2 to -&pi/2:

Ey = -&lambda;0/4&pi;&epsilon;0r&int;cos2 &theta; d&theta; = -&lambda;0/8&epsilon;0r

the force on the charge is the Fy = qEy

we also need to solve for &lambda;0.

Q = &int;dq = &int;&lambda;0rcos &theta; d&theta; = 2r&lambda;0

&lambda;0 = Q/2r

finally
Fy = -qQ/16&epsilon;0r2

plug in the numbers and you get -0.70588 N.