Work to bring a charge to the center of two quarter circles

AI Thread Summary
The discussion revolves around calculating the work required to bring a charge to the center of two quarter circles with opposite linear charge densities. The electric field at point P is derived, leading to the conclusion that the work done is zero due to the condition that the linear charge densities are equal in magnitude but opposite in sign. An intuitive explanation provided indicates that as the charge is brought in along the line y = -x, the electric force acts perpendicular to the motion, resulting in no mechanical work being done. The participants also emphasize the importance of visualizing the arc arrangement for clarity. Overall, the conclusion that the work is zero is confirmed through both mathematical reasoning and intuitive understanding.
lorenz0
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Homework Statement
Find the electric field at point C, which corresponds to the center of the two arcs of circumference with radius ##𝑅 = 10 cm## with uniform charge densities ##\lambda_1 = + 1nC / m## and ##\lambda_2 = -1 nC / m## respectively.
Also find the work required to bring a charge ##q= 5 \mu C## from infinity to point C.
Relevant Equations
##\vec{E}=\frac{kq}{r^2},\ V(r)=\frac{kq}{r}##
By measuring angle \theta from the positive ##x## axis counterclockwise as usual, I get ##d\vec{E}=k( (\lambda_2-\lambda_1)\cos(\theta)d\theta, (\lambda_2-\lambda_1)\sin(\theta)d\theta )## and by integrating from ##\theta=0## to ##\theta=\frac{\pi}{2}## I get ##\vec{E}(P)=\frac{k(\lambda_1-\lambda_2)}{R}(-1,-1)##.

Now, the work to bring a charge from infinity to point P should be (if we set ##V(\infty)=0##) ##L=qV(P)=q\int_{\theta=0}^{\theta=\pi/2}(\frac{k\lambda_1}{R}+\frac{k\lambda_2}{R})Rd\theta=0##

I am a bit unsure about the work being ##0##, it doesn't feel intuitive to me that it should be: is this correct? Is there an intuitive explanation for the work being ##0##? Thanks
 

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Please post either a diagram or a full description of the arc arrangement.
But just looking at the last line, I do not see how you get zero from that integral.
 
haruspex said:
Please post either a diagram or a full description of the arc arrangement.
But just looking at the last line, I do not see how you get zero from that integral.
I have posted the diagram; I get zero since ##\lambda_1=-\lambda_2##.
 
Last edited:
lorenz0 said:
Is there an intuitive explanation for the work being 0?
If you bring in the charge from infinity over the line ##y = -x##, the electric force is perpendicular to the motion, so no mechanical work is involved.

##\ ##
 
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BvU said:
If you bring in the charge from infinity over the line ##y = -x##, the electric force is perpendicular to the motion, so no mechanical work is involved.

##\ ##
I see, thank you very much!
 
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