Electrostatics Question 717
Question: A particle of charge q and mass m moves rectilinearly under the action of electric field $ E=A-Bx, $ where A and B are positive constants and x is distance from the point where particle was initially at rest then the distance traveled by the particle before coming to rest and acceleration of particle at that moment are respectively:
Options:
A) $ \frac{2A}{B},0 $
B) $ 0,-\frac{qA}{m} $
C) $ \frac{2A}{B},-\frac{qA}{m} $
D) $ \frac{-2A}{B},-\frac{qA}{m} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ F=qE=q( A-Bx ) $
$ ma=q( A-Bx )\Rightarrow a=\frac{q}{m}( A-Bx ) $ .(i)
$ \frac{vdv}{dx}=\frac{q}{m}( A-Bx );vdv=\frac{q}{m}q( A-Bx )dx $
$ \int\limits _{0}^{0}{vdv=\frac{q}{m}}\int\limits _{0}^{x}{(A-Bx)dx};Ax-\frac{Bx^{2}}{2}=0 $
$ x=0,x=\frac{2A}{B} $ -(ii)
From eqs. (i) and (ii) $ \frac{q}{m}(A-Bx)=\frac{q}{m}( A-B\times \frac{2A}{B} )=\frac{-aA}{m} $
