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A particle of charge q and mass mmoves rectilinearly under the action of an electric field E = alpha-betax. Here, alpha and beta are positive constants and x is the distance from the point where the particle was initially at rest. Then, the |
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Answer» motion of the particle is oscillatory `a=F/m = (qE)/m=q/m(alpha-betax)` ….(i) a=0 at x=`alpha/beta` i.e., force on the particle is ZERO at x= `alpha/beta` So, mean position of the particle is at x= `alpha/beta` `v(dv)/(DX)=q/m(alpha-betax)` `therefore int_0^v vdv=q/mint_0^x (alpha-betax)dx "" therefore v=sqrt((2qx)/m(alpha-beta/2x))` v=0 at x=0 and `x=(2ALPHA)/beta` So, the particle oscillates between x=0 and `x=(2alpha)/beta` with mean position at `x=alpha/beta` Maximum acceleration of the particle is at extreme position (at x=0 or `x=2alpha//beta` and `a_"max"=qalpha//m` [ from equation (i)] |
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