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How to use the gradient to find Electric field

  1. Mar 8, 2009 #1
    1. A rod carrying a uniform charge distribution is bent into a semi circle with the center on the orgin and a radius R. Calcualte the Electric field at the center of the semi circle using the electric potential expression found in part a



    2. E = -(gradient)V



    3. The electric potential at the center is V = kQ/r or k (pi) (lambda) where lambda is the charge density. The correct expression for the electric field at the center of the circle is 2k(lambda)/r. However, I'm finding that simply taking the gradient of the electric potential function gives kQ/r^2 or k(lambda)(pi)/r, which is not correct. I'm afraid I'm making a fundamental mistake with the definition of the gradient and how to account for the changing electric field direction at the center of the semi circle.
     
  2. jcsd
  3. Mar 9, 2009 #2

    rl.bhat

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    Homework Helper

    Here you have taken the gradient along the radius.
    First you have to finds out delta Ex( due to a small element dl) = -dV/dx and dEy = - dv/dy where dx = dr*cos (theta) and dy = dr*sin(theta). Take summation over the whole semicircular rod to find the field.
     
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