Imperfect Symmetry

Charge is unevenly distributed over a unit circle centered on the origin ( 0 , 0 ) (0,0) . The linear charge density is σ = sin ( θ ) \sigma=\sin(\theta) .

The magnitude of the electric field at the origin can be expressed as α \alpha k k , where k k is the Coulomb constant. Enter the value of α \alpha , given to two decimal places.

Details and Assumptions:

Assume standard SI units for all quantities.

Note that sin ( θ ) \sin(\theta) is negative for some values of θ \theta .


The answer is 3.14.

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1 solution

Steven Chase
Aug 4, 2016

Interesting. One way to improve your solution would be to explain why the fields do not cancel out because of symmetry.

Agnishom Chattopadhyay - 4 years, 10 months ago

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Good point. The top half of the circle is positively charged, and the bottom half is negatively charged. So we're essentially measuring the field in the middle of an electric dipole.

Steven Chase - 4 years, 10 months ago

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So, shouldn't the field vanish because of symmetry? I do not get this, actually.

Agnishom Chattopadhyay - 4 years, 10 months ago

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@Agnishom Chattopadhyay We have perfect symmetry with respect to the vertical axis, because (from a charge distribution point of view), sin(theta) equals sin(pi - theta). Thus the horizontal force vanishes. But since sin(theta) equals -sin(-theta), the vertical picture resembles an electric dipole. Thus, there is a nonzero net field at the center, which is purely vertical.

Steven Chase - 4 years, 10 months ago

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@Steven Chase Thanks, that was helpful.

Agnishom Chattopadhyay - 4 years, 10 months ago

Shouldn't r \vec{r} be r = cos θ i ^ sin θ j ^ \vec{r}=-\cos \theta \hat{i}-\sin \theta \hat{j} ?

Alan Enrique Ontiveros Salazar - 4 years, 10 months ago

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Yes, you're right. But I didn't worry about it since I was only looking for the magnitude.

Steven Chase - 4 years, 10 months ago

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Awesome question by @Steven Chase sir and also Awesome solution by @Steven chase sir. GOOD problem. Thanks

A Former Brilliant Member - 1 year, 7 months ago

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