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 08-25-2005, 07:00 PM #1 (permalink) Banned   Join Date: Aug 2005 Posts: 16 Physics Help Hey guys I know this is the Homework forum so I need help on my physics. A half-ring with a radius 8 cm has a total charge 3 ÂµC uniformly distributed along its length. A half-ring with a radius 8 cm has a total charge 3 ÂµC uniformly distributed along its length. I found the charge density on the ring to be .0000119 (1.19 e -15 C/M). Now I'm trying to set up my integral to figure out the total Ex since that will be Enet since Ey's cancel. I know that for a small charge dQ on the ring, the E field produced is kdQ/r^2 and the Ex component would be k(dQ)/r^2cos(theta). So now I'm trying to define dQ and to figure out an appropriate integral to find the total E field. I tried dQ=Lamda * dx (which in hindsight didn't make a lot of sense). So i set the integral to int( (k*lambda/r^2) * cos(theta) dx ). I then converted cos(theta) to x/r and moved all the non variables out front so I just have (k*lambda/r^3)*int(xdx). I then set the limits of the integral to 0, .08 since the minimum x distance is 0 when looking at the top of the ring and X=R when directly to the left. I then multiplied the whole thing by 2 to account for the bottom half. When I get the answer it's half of what I should be getting. Where am I going wrong? What should be the appropriate definition of dQ in this case? __________________
08-25-2005, 07:13 PM   #2 (permalink)
Wizard Techie

Join Date: Feb 2005
Location: Milford, Connecticut
Posts: 4,218
Re: Physics Help

Quote:
 Originally posted by BOYAKASHA Hey guys I know this is the Homework forum so I need help on my physics. A half-ring with a radius 8 cm has a total charge 3 ÂµC uniformly distributed along its length. A half-ring with a radius 8 cm has a total charge 3 ÂµC uniformly distributed along its length. I found the charge density on the ring to be .0000119 (1.19 e -15 C/M). Now I'm trying to set up my integral to figure out the total Ex since that will be Enet since Ey's cancel. I know that for a small charge dQ on the ring, the E field produced is kdQ/r^2 and the Ex component would be k(dQ)/r^2cos(theta). So now I'm trying to define dQ and to figure out an appropriate integral to find the total E field. I tried dQ=Lamda * dx (which in hindsight didn't make a lot of sense). So i set the integral to int( (k*lambda/r^2) * cos(theta) dx ). I then converted cos(theta) to x/r and moved all the non variables out front so I just have (k*lambda/r^3)*int(xdx). I then set the limits of the integral to 0, .08 since the minimum x distance is 0 when looking at the top of the ring and X=R when directly to the left. I then multiplied the whole thing by 2 to account for the bottom half. When I get the answer it's half of what I should be getting. Where am I going wrong? What should be the appropriate definition of dQ in this case?
Wrong forum. Here may-be this would help!
http://www.physicsforums.com/
__________________

 08-25-2005, 07:27 PM #3 (permalink) Monster Techie   Join Date: Apr 2005 Posts: 1,107 Computer HW = Hard Ware __________________ Check out www.advancedgeeks.com Great new forum!
08-25-2005, 07:37 PM   #4 (permalink)
Monster Techie

Join Date: Jan 2005
Posts: 1,523
Re: Physics Help

Quote:
 Originally posted by BOYAKASHA When I get the answer it's half of what I should be getting. Where am I going wrong?
Yeah... I can see why this would concern you. CLEARLY, you missed a factor of 2 in your equations. So multiply your answer by 2 and you should get the correct answer. Hope this helped.

Yes.. It was a stupid answer.. THat's all I give a SPAMMER!!

 08-25-2005, 07:39 PM #5 (permalink) Monster Techie   Join Date: Apr 2005 Posts: 1,107 Lol Chank once again saves t3h day -.- :-P __________________ Check out www.advancedgeeks.com Great new forum!
08-26-2005, 12:22 AM   #6 (permalink)
Banned

Join Date: Jul 2005
Posts: 2,327
Re: Re: Physics Help

Quote:
 Originally posted by Chankama Yeah... I can see why this would concern you. CLEARLY, you missed a factor of 2 in your equations. So multiply your answer by 2 and you should get the correct answer. Hope this helped. Yes.. It was a stupid answer.. THat's all I give a SPAMMER!!
I have to admit, that was good, i'm impressed.
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