# Uniform charge Electric field

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

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

Thank you.

Stephen

LowlyPion
Homework Helper
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.

Anybody have any opinions.

I really appreciate the help.
Stephen

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

Any help would be appreciated. Thanks.
Stephen

Redbelly98
Staff Emeritus