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

  • \$\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
    Commented Oct 23, 2014 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
    Commented Oct 24, 2014 at 7:02

1 Answer 1


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.

  • \$\begingroup\$ But other points also add to the field. \$\endgroup\$
    – Desperado
    Commented Oct 24, 2014 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
    Commented Oct 26, 2014 at 9:18

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