Electric field

Asked by angkit | 30th May, 2010, 09:14: PM

Expert Answer:

Dear student,

The electric field at the center of a semicircle loop of radius R carrying uniform charge Q distributed uniformly over its length  (Fig. 6).

  1. Take symmetrically distributed elements 1 and 2 of length ds and charge dq = lds.  Since dQ = ds/R in radians, dq = lRdQ.

  2. dE1 = dE2 = kdq/R2 = dE.
    The directions are as shown in Fig. 6.
    The X-components of dE1 and dE2 cancel.
    dEy = dE sin Q = (kdq/R2) sin Q = klRdQsin Q/R2 = kldQsin Q/R.
  3. E = 
  4. Since l= charge/length = Q/pR,
    E = 2k(Q/pR)R = 2(1/4peo)(Q/pR2)= Q/2 p2eoR2.
    E = - Q/2p2eoR2 j
  5.  

    Answered by  | 31st May, 2010, 11:33: AM

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