Semicircular Wire (Couloumb's law)

[Fabio010]

Semicircular Wire
A positively charged wire is bent into a semicircle of radius R, as shown in Figure 2.15.4.

The total charge on the semicircle is Q. However, the charge per unit length along the
semicircle is non-uniform and given by

λ = λo θ cos .

(a) What is the relationship betweenλo , R and Q?

(b) If a charge q is placed at the origin, what is the total force on the charge?



Attempts:

a)

dq = λdl = λRdθ

dq = λo.cosθ.R.dθ

Q = ∫(-pi/2 to pi/2) [λo.cosθ.R.dθ]

Q = 2λoR



b)

Force in x-axis = ((Ke.q.λo.cosθ.R.dθ)/(R^2))*sinθ integrating that in order to dθ, the result is zero.

Force in y-axis = ((Ke.q.λo.cosθ.R.dθ)/(R^2))*cosθ integrating that in order to dθ, the result is (Ke.q.λo/R)*(0.5*pi)



Is that correct??

 

[TSny]

Bingo! Good work.

 

[Chestermiller]

To make your professor happy, you should solve for λ₀ in terms of Q, and substitute that into your final answer.

 

[Fabio010]

To make your professor happy, you should solve for λ₀ in terms of Q, and substitute that into your final answer.

lol ok thanks for the help!

 

[paranormal]

Your attempts are mostly correct. However, in part a), the integral should be from 0 to pi, not from -pi/2 to pi/2. This is because the semicircle only covers the positive values of θ.

Also, for part b), the force in the x-axis should be integrated from 0 to pi/2, and the force in the y-axis should be integrated from pi/2 to pi. This is because the force changes direction at the point where the semicircular wire changes direction.

Therefore, the final equations for the forces would be:

Force in x-axis = ((Ke.q.λo.cosθ.R.dθ)/(R^2))*sinθ integrating from 0 to pi/2

Force in y-axis = ((Ke.q.λo.cosθ.R.dθ)/(R^2))*cosθ integrating from pi/2 to pi

Overall, your understanding of Coulomb's law and its application to a semicircular wire is correct. Keep up the good work!