Asked by angkit | 30th May, 2010, 09:14: PM
The electric field at the center of a semicircle loop of radius R carrying uniform charge Q distributed uniformly over its length (Fig. 6).
- Take symmetrically distributed elements 1 and 2 of length ds and charge dq = lds. Since dQ = ds/R in radians, dq = lRdQ.
- 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.
- E =
- Since l= charge/length = Q/pR,
E = 2k(Q/pR)R = 2(1/4peo)(Q/pR2)= Q/2 p2eoR2.
E = - Q/2p2eoR2 j
Answered by | 31st May, 2010, 11:33: AM
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