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Electric Field Calculations for Uniform Ring of Charge and Uniformly Charged Disk. Montwood High School AP Physics C R. Casao. Electric Field of a Uniform Ring of Charge. Montwood High School AP Physics C R. Casao. Consider the ring as a line of charge that has been formed into a ring.

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Electric field calculations for uniform ring of charge and uniformly charged disk

Electric Field Calculations for Uniform Ring of Charge and Uniformly Charged Disk

Montwood High School

AP Physics C

R. Casao


Electric field of a uniform ring of charge

Electric Field of a Uniform Ring of Charge Uniformly Charged Disk

Montwood High School

AP Physics C

R. Casao


  • Consider the ring as a line of charge that has been formed into a ring.

    • Divide the ring into equal elements of

      charge dq; each element of charge dq

      is the same distance r from point P.

    • Each element of charge dq can be considered as a point charge which

      contributes to the net electric field at

      point P.


  • At point P, the electric field contribution from each element of charge dq can be resolved into an x component (Ex) and a y component (Ey).

  • The Ey component for the electric field from an element of charge dq on one side of the ring is equal in magnitude but opposite in direction to the Ey component for the electric field produced by the element of charge dq on the opposite side of the ring (180º away). These Ey components cancel each other.



  • cos element of charge dq can be resolved into an x component (Eq can be expressed in terms of x and r:

  • The total electric field can be found by adding the x-components of the electric field produced by each element of charge dq.

  • Integrate around the circumference of the ring:


  • is the symbol for integrating around element of charge dq can be resolved into an x component (E

    a closed surface.

  • Left side of the integral: adding up all the little pieces of dEx around the circumference gives us Ex (the total electric field at the point).

  • Right side of the integral: pull the constants k, x, and r out in front of the integral sign.




Mit visualizations
MIT Visualizations ring, a, and the position on the x-axis, x.

  • URL: http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/electrostatics/index.htm

  • The Charged Ring

  • Integrating Around a Ring of Charge


Electric field of a uniformly charged disk

Electric Field of a Uniformly Charged Disk ring, a, and the position on the x-axis, x.


  • Surface charge density: ring, a, and the position on the x-axis, x.

  • Divide the disk into concentric rings which will increase in size from the center of the disk to the outer rim of the disk.

  • r is the distance from the center of the disk to a particular ring.

  • Each ring will have a different charge, radius, and area.



  • For each ring: disk to the ring location, so does the amount of charge on the ring and the area of the ring.


  • dq is expressed in terms of dr because the radius of each ring will vary from the center of the disk to the rim of the disk.

  • The charge within each ring can be divided into equal elements of charge dq, which can then be treated as point charges which contribute to the electric field at point P (see the ring problem).

  • Point charge equation:





  • Express the cos the x-components contribute to the net electric field at point P.q in terms of the variables x and r. L is the distance from dq to point P.




  • Substitution method: together gives us the net electric field, E

    • Let u = r2 + x2

    • Then du = 2·r dr + 0; du = 2·r dr.

    • The derivative of x2 is 0 because it is a constant and the derivative of a constant is 0; r is a quantity that changes.



  • So: together gives us the net electric field, E



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