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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