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Help with electromagnetism

  • Thread starter mateoguapo327
  • Start date
  • #1
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.

the answer in the book is -0.707Nj and I got -0.526N. Can someone please help me figure out how to properly go about solving this problem?
 

Answers and Replies

  • #2
653
0


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:

dEy = - 1/4πε0 cos θ dq/r2 = - 1/4πε0 cos θ λrdθ/r2 = - 1/4πε0 cos2 θλ0dθ/r

then integrate from π/2 to -&pi/2:

Ey = -λ0/4πε0r∫cos2 θ dθ = -λ0/8ε0r


the force on the charge is the Fy = qEy


we also need to solve for λ0.

Q = ∫dq = ∫λ0rcos θ dθ = 2rλ0

λ0 = Q/2r

finally
Fy = -qQ/16ε0r2


plug in the numbers and you get -0.70588 N.
 
  • #4
653
0
oops. my bad.
 

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