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Phymath
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A uniformly charged infinitely thin circular ring of radius a has total charge Q. Place the ring in the xy-plane with its center at the origin. Use cylindrical coordinates...
Find the potential at a point in the xy-plane a distance 2a from the orgin give your answer as a number times kQ/a
...
I figured the potential to be for some point in the xy-plane,[tex] (2a, \Phi , 0)[/tex]
[tex]k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{(2a)^2+a^2-2(2a)a cos(\Phi - \theta)}}[/tex]
[tex]= k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{4a^2+a^2-4a^2 cos(\Phi - \theta)}}[/tex]
where, [tex] j = \frac{Q}{2 \pi a}[/tex]
[tex]= k_e j \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{5-4cos(\Phi - \theta)}}[/tex]
did I set up this intergral right? how do I evaluate it? hints answers it all helps
Find the potential at a point in the xy-plane a distance 2a from the orgin give your answer as a number times kQ/a
...
I figured the potential to be for some point in the xy-plane,[tex] (2a, \Phi , 0)[/tex]
[tex]k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{(2a)^2+a^2-2(2a)a cos(\Phi - \theta)}}[/tex]
[tex]= k_e j a \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{4a^2+a^2-4a^2 cos(\Phi - \theta)}}[/tex]
where, [tex] j = \frac{Q}{2 \pi a}[/tex]
[tex]= k_e j \int_0 ^{2 \pi} \frac{d \theta}{\sqrt{5-4cos(\Phi - \theta)}}[/tex]
did I set up this intergral right? how do I evaluate it? hints answers it all helps
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