Electric field E =- bx + a exists in a region parallel to the x-direction ( a and b are positive constants). A charge particle having charge q and mas s m is released from the origin x = 0. If a = 0.50 N /C and b = 0.20 N/Cm, and qm=50C/kg find the magnitude o f the acceleration of the particle (in m/s2) at the instant its speed becomes zero for the first time after release.

The force on the charged particle, F=qE=q(−bx+a)ma=q(−bx+a)[a= acceleration ]dvdx=−bqmx+aqm∫0v vdv=−bqm∫0x xdx+aqm∫0x dxv22=−bqx22m+aqxmWhen speed of the particle becomes zero, v = 0 when bqx22m=aqxm⇒x=2abAs the acceleration o f the particle is given as a0=qm(−bx+a)  ………(i)The acceleration at x=2abHence the acceleration =qm−b⋅2ab+a=qm(−2a+a)=−qam                                       =−50×0.5=−25m/s2