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StephenDoty
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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
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