0
\$\begingroup\$

So, there's a tape with width of 2a, -a on the -y and a on the y, and it very long, penetrating the the plane of picture, it's charge density (dQ/dS) is

σ=σ0 * y /a , σ0 is a constant.

What is the electric field in point A(a.0)?

I tried solving this using charged disk, expressing electric field over polar coordinates etc., that is when there's a uniform charge. But I have no idea how to express y using shapes. What to do?

this axis

\$\endgroup\$
2
  • \$\begingroup\$ What is "A(a.0)"? Where is it? Your graph is just an empty graph of x and y - is there any relevance to this empty graph? \$\endgroup\$
    – Andy aka
    Oct 23 '14 at 22:54
  • \$\begingroup\$ A point with those coordinates, it's a half of the width on the x axis. I wanted to help you visualize. You can't see the length of the tape, because it's penetrating the plane of the system, it's very thin so nothing on x, and the only dimensions are from a to -a, on y axis. \$\endgroup\$
    – Desperado
    Oct 24 '14 at 7:02
1
\$\begingroup\$

I think you must use the superposition principle of the field in this case, and consider the problem as a distribution of charge along a line from a to -a, because the contribution to the field of the charge along the z axe is 0, the field in this case is always parallel to the surface xy. The problem is to build the right integral considering a dE and a dL (along the y axe). enter image description here

Note: there is a little mistake, I put r=x^2+y^2 instead r^2=x^2+y^2, I have forgot to put the ^2 on the r, but I have taken into account in the rest of the calculation.

\$\endgroup\$
2
  • \$\begingroup\$ But other points also add to the field. \$\endgroup\$
    – Desperado
    Oct 24 '14 at 7:06
  • \$\begingroup\$ @Desperado: I have updated the answer, now I think it is pointing to the rigth way to solve the problem. \$\endgroup\$
    – Alf
    Oct 26 '14 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.