text: Find the force experienced by the s...
  Question
  
  # Find the force experienced by the semicircular rod charged with a charge q, placed as shown in figure. Radius of the wire is ( mathrm{R} ) and the infinitely long line of charge with linear density ( lambda ) is passing through its to the plane of wirn centre and perpendicular to the plane of wire.(A) ( frac{lambda q}{2 pi^{2} varepsilon_{0} R} )(B) ( frac{lambda q}{pi^{2} varepsilon_{0} R} )(C) ( frac{lambda q}{4 pi^{2} varepsilon_{0} R} )(D) ( frac{lambda q}{4 pi varepsilon_{0} R} )
  
  JEE/Engineering Exams
  Physics
  Solution
  64
  4.0 (1 ratings)
  
  ( d F=E cos theta d q( ) where Eistheelectricfield
  ofthechargedrod Electricfieldatthelocationofthechargedq
  ( i s )
  ( E=frac{lambda}{2 pi in R} )
  ( d q=frac{q}{pi R} times R d theta )
  ( d F=frac{lambda cos theta}{2 pi in R} frac{q}{pi R} times R d theta )
  ( int d F=int frac{lambda cos theta}{2 pi epsilon R} frac{q}{pi} times d theta )
  ( F=int_{-frac{pi}{2}}^{+frac{pi}{2}} frac{lambda cos theta}{2 pi in R} frac{q}{pi} times d theta )
  ( F=int_{-frac{pi}{2}}^{+frac{pi}{2}} frac{lambda cos theta}{2 pi in R} frac{q}{pi} times d theta )
  ( F=frac{lambda q}{2 pi^{2} in R}(sin theta)^{frac{pi}{2}}=frac{2 lambda q}{2 pi^{2} in R} )
  ( F=frac{lambda q}{pi^{2} in R} )