Uniform charge Electric field

[StephenDoty]

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*$$\sqrt{r^2 +(L/2)^2}$$)

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

From glass rod
E at 1cm = E= k 12nC/((.01m)*$$\sqrt{(.01m)^2 +(.05m)^2}$$)= 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)*$$\sqrt{(.029m)^2 +(.05m)^2}$$) = 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

 

[StephenDoty]

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

Thank you.

Stephen

 

[LowlyPion]

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.

 

[StephenDoty]

Anybody have any opinions.

I really appreciate the help.
Stephen

 

[StephenDoty]

I have to turn this in very soon. Did I do this problem correctly?

Any help would be appreciated. Thanks.
Stephen

 

[Redbelly98]

Yes, it looks good. You might watch the significant figures, if your professor cares about that sort of thing.