Potential for semi-circle

[brad sue]

I have this problem for homework and I would like to check my work:

Charges are distribyted with uniform charge density (lambda)along a semi circle of radius R, centered at the origin of a coordinate system.
What is the potential at the origin?

I did this:
dV=k*dq/R where k=1/4*pi*(epsilon 0)


but dq=(lambda)*R*d(theta) theta is the small angle between two position on the semi circle.

pi
then V =| k*(lambda)*R*d(theta) / R (integral from 0 to pi)
0
After computation I found: V=(lambda)/(4*epsilon 0)

Thank you

 

[Galileo]

Looks ok.
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[kyle8921]

for sharing your work and question. I believe your approach to solving this problem is correct. By using the equation for potential (V=k*q/r), where k is the Coulomb constant and q is the charge, and integrating over the semi circle with a small angle d(theta), you can find the potential at the origin. This is a good application of the concept of charge density, which represents the amount of charge per unit area. By considering the charge density along the semi circle, you were able to find the total charge and use it in the potential equation. Your final answer of V=(lambda)/(4*epsilon 0) is correct. Keep up the good work!