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?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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?

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Do i use \[kQ \div r \sqrt{r ^{2}+(L/2)^{2}}\] ? i use this but then i dont know what to do next
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.)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question