Finding an expression for Electric Potential?

[PolarBee]

Homework Statement

There is a thin rod with charge Q that has been bent into a semicircle with radius R. Find an expression for the electric potential at the center.

Radius = R
Charge = Q

Homework Equations

V = ∫(k * dq)/r

The Attempt at a Solution

dq = λdx
λ = Q/L
L = pi * r

V = kQ / LR ∫ dx = kQ / pi * r^2 ∫ [3pi/2 - pi/2]
V = kQ/R^2

But the answer should be kQ/R. How did the other R get cancelled?

 

[kuruman]

What are your limits of integration? You should use polar coordinates and integrate over an angle.

 

[PolarBee]

I integrated from pi/2 to 3pi/2. Which came out to be just pi and that canceled out the pi from L = pi * r.

 

[kuruman]

I integrated from pi/2 to 3pi/2. Which came out to be just pi and that canceled out the pi from L = pi * r.

Your integrand has a dx in it. What did you do with it? To integrate over an angle you need a dθ in the integrand.

 

[PolarBee]

Oh I see. So dx should be rdθ instead for polar coordinates right, without it I can't integrate from pi/2 to 3pi/2?

 

[kuruman]

Correct. That cancels the extra R in the denominator. Also, note that the beginning angle is arbitrary. A semicircle is a semicircle. You might as well integrate from 0 to π.