A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
A 10cmlong thin glass rod uniformly charged to 10.0 nC and a 10cmlong 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?
anonymous
 4 years ago
A 10cmlong thin glass rod uniformly charged to 10.0 nC and a 10cmlong 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?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do i use \[kQ \div r \sqrt{r ^{2}+(L/2)^{2}}\] ? i use this but then i dont know what to do next

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1At the risk of messing up again:dw:1329412984230:dwfor each point call the distance from the left rod r1 and from the right rod r2=0.038r1. The charges are opposite so the fields will add\[\sum E=E_1+E_2=\frac{kQ}{r_1\sqrt{r_1^2+(\frac L2)^2}}+\frac{kQ}{r_2\sqrt{r_2^2+(\frac L2)^2}}\]\[=kQ(\frac1{r_1\sqrt{r_1^2+(\frac L2)^2}}+\frac1{(0.038r_1)\sqrt{(0.038r_1)^2+(\frac L2)^2}})\]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.)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.