Finding an expression for Electric Potential?

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Homework Help Overview

The discussion revolves around finding an expression for the electric potential at the center of a semicircular rod with charge Q and radius R. The original poster attempts to derive the potential using integration but encounters a discrepancy in their result.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the limits of integration and the appropriate use of polar coordinates. There is a focus on the correct form of the differential element in the integration process.

Discussion Status

Some participants have offered guidance on the integration approach, suggesting the need to use dθ instead of dx for polar coordinates. The conversation reflects a productive exploration of the problem, with participants questioning assumptions and clarifying the integration method.

Contextual Notes

There is a mention of the arbitrary nature of the starting angle for the semicircle, indicating that different approaches to setting limits may yield similar results. The original poster's confusion about the cancellation of terms in their expression is also noted.

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

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