How to Calculate Total Force on a Charge at the Center of Curvature?

[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?

 

[lethe]

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_y = - 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:

E_y = -λ₀/4πε₀r∫cos² θ dθ = -λ₀/8ε₀r


the force on the charge is the F_y = qE_y


we also need to solve for λ₀.

Q = ∫dq = ∫λ₀rcos θ dθ = 2rλ₀

λ₀ = Q/2r

finally
F_y = -qQ/16ε₀r²


plug in the numbers and you get -0.70588 N.

 

[quantumdude]

Ahem!

Mateoguapo:

https://www.physicsforums.com/showthread.php?s=&threadid=28

In the future, please show your work first.

Thanks,

 

[lethe]

oops. my bad.