What is the electric potential at the center of a charged half-circular washer?

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Homework Help Overview

The problem involves calculating the electric potential at the center of a charged half-circular washer with a uniform charge distribution. The inner and outer radii of the washer are defined as 'a' and 'b', respectively, and the potential at infinity is set to zero.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of electric field equations and potential equations, with one participant attempting to derive the potential using integration. Others question the correctness of the algebra and the treatment of electric fields as vectors rather than scalars.

Discussion Status

Some participants have provided guidance on focusing on potentials instead of fields, suggesting that potentials can be added directly. There are ongoing questions about the integration limits and methods for integrating in polar coordinates.

Contextual Notes

Participants express confusion regarding the integration process and the application of formulas, indicating a need for clarification on integrating over an area in polar coordinates.

annastewert
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Homework Statement


A plastic circular washer is cut in half and has a charge Q spread uniformly over it. If the electrical potential at infinity is taken to be zero, what is the electric potential at the point P, the center of the old washer? The inner radius of the washer is a, the outer radius is b.
*see attached picture*

Homework Equations


I know that you can use ∫∫E⋅dA=Q/ε and solve for E. Then V(r)=∫E(r)dr

The Attempt at a Solution


∫∫E⋅dA=Q/ε
E∫∫dA=Q/ε
∫∫dA=πb^2-πa^2
E(πb^2-πa^2)=Q/ε
E=Q/ε(πb^2-πa^2)

Then I integrated this, but the answer is V=Q/2πε(b+a) which is not what I am getting. Where am I going wrong?
 

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Your algebra is wrong for two reasons. First, the field at P due to element dA depends on the position of the element dA in relation to P. You took it outside the integral as though it is constant. Secondly, E is a vector, not a scalar. An integral is a sum, and vectors add differently to the way scalars add. The field at P due to different elements will point in different directions, so when you add them there will be cancellation.

The easiest way to solve this problem is to start with potentials, not fields. Potentials also add. What is the potential at P due to an element dA at location (r,θ) in polar coordinates centred on P? What is dA equal to in terms of dr and dθ?
 
I'm confused on how to do this just with potentials, do you use the equation V=Q/(4πεr) and then integrate from a to b? but you would also need to integrate from 0-π wouldn't you? How would you do this?
 
annastewert said:
I'm confused on how to do this just with potentials, do you use the equation V=Q/(4πεr) and then integrate from a to b? but you would also need to integrate from 0-π wouldn't you? How would you do this?
Do you know how to integrate over an area in polar coordinates?
 
I don't think so no.
 
annastewert said:
I don't think so no.
Given a function f(r,θ), the integral over an area A is ∫Af r dr dθ.
 

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