Uniform charge Electric field

  • #1
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A 10cm long thin glass rod uniformly charged to 12nC and a 10cm long thin plastic rod uniformly charged to -12nC are placed side by side, 3.9cm apart. What are the electric field strengths at 1cm, 2 cm , and 3cm from the glass rod along the line connecting the midpoints of the two rods?

E=k Q/(r*[tex]\sqrt{r^2 +(L/2)^2}[/tex])

E=k Q/(r*[tex]\sqrt{r^2 +(.05m)^2}[/tex])

From glass rod
E at 1cm = E= k 12nC/((.01m)*[tex]\sqrt{(.01m)^2 +(.05m)^2}[/tex])= 211805 N/C
E at 2cm = 100275 N/C
E at 3cm = 61739.5 N/C

From plastic rod
E at (3.9cm - 1cm) = E= k 12nC/((.029m)*[tex]\sqrt{(.029m)^2 +(.05m)^2}[/tex]) = 64429.9 N/C
E at (3.9cm - 2cm) = 106270 N/C
E at (3.9cm - 3cm) = 236204 N/C

E net at p1= 211805 N/C + 64429.9 N/C = 276235 N/C
E net at p2= 100275 N/C + 106270 N/C = 206545 N/C
E net at p3 = 61739.5 N/C + 236204 N/C = 297944 N/C

Did I do this right?

Thanks for the help.
Stephen
 

Answers and Replies

  • #2
265
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I am really unsure if I did this problem correctly. Any advise would be greatly appreciated.

Thank you.

Stephen
 
  • #3
LowlyPion
Homework Helper
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I am really unsure if I did this problem correctly. Any advise would be greatly appreciated.

Thank you.

Stephen
Without looking up the charged rod relationship or the math involved, the method looks good.
 
  • #4
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Anybody have any opinions.

I really appreciate the help.
Stephen
 
  • #5
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I have to turn this in very soon. Did I do this problem correctly?

Any help would be appreciated. Thanks.
Stephen
 
  • #6
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,116
152
Yes, it looks good. You might watch the significant figures, if your professor cares about that sort of thing.
 

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