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} ) Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free