would anyone like to attemt to solve this problem? no one at PF seems interested in attempting it. http://www.physicsforums.com/showthread.php?threadid=1415 my friend would appreciate you help..... if not, i'll just have to ask my physics professor i guess thanks
The 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 <font face="symbol">l</font>= <font face="symbol">l</font><sub>0</sub> cos <font face="symbol">q</font>. 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. is the angle formed by dragging R clockwise from the positive y axis. the answer in the book is -0.707N<b>j</b> and I got -0.526N. Can someone please help me figure out how to properly go about solving this problem?
First, DRAW A PICTURE! Then, using symmetry and a little calculus, you can easily solve this problem: First, note that charge density is symmetric w.r.t. the left and right halves of the semicircle; this allows you to neglect any force in the X-direction because for every little charge element, dq, on the right half, there is an equivalent one on the upper half. (This explains the J unit vector). Now, just focus on the force in the y-direction that is contributed by each little "piece" of charge that make up the hoop. The charge of each "piece" is equal to dq=(radius)x(dTheta)x((lambda)xCos(theta)). The force in the negative Y direction for each of these "pieces" is the constant, k, (or one over four*pi*epsilon naught) times (dq*Q/r^2)xCos(theta), where Q is the charge in the center. From here, you just need to integrate the force from pi/2 to negative pi/2. You will also probably have to explain the symmetry arguments to get full credit; you could just do an integral for the X-force, but it would be zero. Therefore, just do a little explaining and save yourself the trouble. If you still get the wrong answer, book might be wrong... I didn't check your results. Good luck
repost from pf consider the electric field at the center of the semicircle due to a little piece of the semicircle at angle θ. by symmetry, only the vertical component of this field contributes: dE<sub>y</sub> = - 1/4πε<sub>0</sub> cos θ dq/r<sup>2</sup> = - 1/4πε<sub>0</sub> cos θ λrdθ/r<sup>2</sup> = - 1/4πε<sub>0</sub> cos<sup>2</sup> θλ<sub>0</sub>dθ/r then integrate from π/2 to -&pi/2: E<sub>y</sub> = -λ<sub>0</sub>/4πε<sub>0</sub>r∫cos<sup>2</sup> θ dθ = -λ<sub>0</sub>/8ε<sub>0</sub>r the force on the charge is the F<sub>y</sub> = qE<sub>y</sub> we also need to solve for λ<sub>0</sub>. Q = ∫dq = ∫λ<sub>0</sub>rcos θ dθ = 2rλ<sub>0</sub> λ<sub>0</sub> = Q/2r finally F<sub>y</sub> = -qQ/16ε<sub>0</sub>r<sup>2</sup> plug in the numbers and you get -0.70588 N.
Aw man. Lethe beat me to it. Please Register or Log in to view the hidden image! Someday I will be able to do that. Right now I am on Chapter 2 of Pre-Calc. I have AT LEAST a few more chapters to go to catch up to him. Please Register or Log in to view the hidden image!
