A) \[\frac{2A}{B},0\]
B) \[0,-\frac{qA}{m}\]
C) \[\frac{2A}{B},-\frac{qA}{m}\]
D) \[\frac{-2A}{B},-\frac{qA}{m}\]
Correct Answer: C
Solution :
[c] \[F=qE=q\left( A-Bx \right)\] \[ma=q\left( A-Bx \right)\Rightarrow a=\frac{q}{m}\left( A-Bx \right)\] ?.(i) \[\frac{vdv}{dx}=\frac{q}{m}\left( A-Bx \right);\,\,\,vdv=\frac{q}{m}q\left( A-Bx \right)dx\] \[\int\limits_{0}^{0}{vdv=\frac{q}{m}}\int\limits_{0}^{x}{(A-Bx)dx}\,\,\,\,;\,\,\,\,Ax-\frac{B{{x}^{2}}}{2}=0\] \[x=0,x=\frac{2A}{B}\] ?(ii) From eqs. (i) and (ii) \[\frac{q}{m}(A-Bx)=\frac{q}{m}\,\left( A-B\times \frac{2A}{B} \right)\,=\frac{-\,aA}{m}\]You need to login to perform this action.
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