InterviewSolution
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InterviewSolution
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An infinite sheet carrying a uniform surface charge density sigma1lies on the xy - plane . The work done to carry a charge q from the point vecA=a(hati-2hatj+6hatk)to the point vecB=a(hati-2hatj+6hatk)(where a is a constant with the dimension of length and epsilon_(0)is the permittivity of free space) is |
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Answer» `(3sigmaaq)/(2epsilon_(0))` DISPLACEMENT VECTOR `vecr` from POINT A to B is `vecr=vecr_(B)-vecr_(A)` `=a(hati-2hatj+6hatk)-a(hati+2hatj+3hatk)=a(-4hatj+3hatk)` Electric field due to an infinite plane sheet of uniformsurface charge density `SIGMA` is `vecE=(sigma)/(2epsilon_(0))hatn`where `hatn`is a unit vector normal to the plane. Here ,`hatn=hatk` `thereforevecE=(sigma)/(2epsilon_(0))hatk` Work done in moving a charge q from A to B is `W=vecF*vecr=qvecE*vecr ""(becausevecF=qvecE)` `=q((sigma)/(2epsilon_(0))hatk)*a(-4hatj+3hatk)=(3sigmaaq)/(2epsilon_(0))` |
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