CBSE Class 12-science Answered

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.
   dE_y = 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(1/4pe_o)(Q/pR²)= Q/2 p²e_oR².
   E = - Q/2p²e_oR² j
5.