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Two semicircle wires ABC, and ADC, each of radius R are lying on xy and xz planes, respectively as shown in fig. if the linear charge density of the semicircle parts and straight parts and straight parts is `lambda`, Find the electric field entensity `vec(E)` at the origin . |
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Answer» Correct Answer - `-(lambda)/(2 pi epsilon_(0)R)(hat j+hat k)`. Electric field due to MA: `vecE_(1)=(lambda)/(4 pi epsilon_(0)R)(hati -hatk)` Electric field due to ADC: `vecE_(2)=(lambda)/(2 pi epsilon_(0)R)(-hatk)` Electric field due to ABC: `vecE_(3)=(lambda)/(2 pi epsilon_(0)R)(-hatj)` Electric field due to NC: `vecE_(4)=(lambda)/(4 pi epsilon_(0)R)(hati-hatk)` Net electric field is `vecE_(0)=vecE_(1)+vecE_(2)+vecE_(3)+vecE_(4)=-(lambda)/(2 pi epsilon_(0)R)(hat j+hat k)`. |
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