• anonymous
A 10-cm-long thin glass rod uniformly charged to 10.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 10.0 nC are placed side by side, 3.80 cm apart. What are the electric field strengths E_1 to E_3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
  • Stacey Warren - Expert
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  • katieb
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  • anonymous
Do i use \[kQ \div r \sqrt{r ^{2}+(L/2)^{2}}\] ? i use this but then i dont know what to do next
  • TuringTest
At the risk of messing up again:|dw:1329412984230:dw|for each point call the distance from the left rod r1 and from the right rod r2=0.038-r1. The charges are opposite so the fields will add\[\sum E=E_1+E_2=\frac{kQ}{r_1\sqrt{r_1^2+(\frac L2)^2}}+\frac{kQ}{r_2\sqrt{r_2^2+(\frac L2)^2}}\]\[=kQ(\frac1{r_1\sqrt{r_1^2+(\frac L2)^2}}+\frac1{(0.038-r_1)\sqrt{(0.038-r_1)^2+(\frac L2)^2}})\]and r1 will be each respective distance from the first rod. (PS:I think perhaps if I had used the formula for the field due to a finite rod we would have gotten your other question right, but I imagined it to be infinite rods last time.)

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