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An infinitely long wire with linear charge density+lambdais bent and the two parts are inclined at angle 30^@ as shown in fig, the electric field at point P is : |
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Answer» `(lambda)/(4 pi in_0 d) [ (1 + sqrt(3))hati + hatj]` `E_1 = (lambda)/(4 pi in_0 ((sqrt(3)d)/(2))) (COS 60^@ + cos 0^@) = (sqrt(3) d)/(4 pi in_0 d)` `:. VEC(E ) = (E_1 + E_2 cos 30^@ + E_2 cos 60^@) hati + (E_1^. + E_2^1 sin 60^@ - E_2 sin 30^@) hati`  `= ((lambda)/(4pi in_0 d) + (sqrt(3)lambda)/(8pi in_0 d) + (sqrt(3) lambda)/(8 pi in_0 d)) hati + ((lambda)/(4pi in_0 d) + (sqrt(3))/(8pi in_0 d) - (sqrt(3))/(8pi in_0 d)) hatj` `= (lambda)/(4 pi in_0 d)[(1 + sqrt(3) hati + 2hatj]` |
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