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Q.

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 mass m is released from the origin x = 0. Find the acceleration of the particle at the instant its speed becomes zero for the first time after release.

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a

$\frac{q b}{m}$

b

$- \frac{q b}{m}$

c

$\frac{q a}{m}$

d

$- \frac{q a}{m}$

answer is D.

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Detailed Solution

$F = q E = q (a - b x) = m a = m v \frac{d v}{d x} \Rightarrow \int_{0}^{v} v . d v = \frac{q}{m} \int_{0}^{x} (a - b x) d x \Rightarrow \frac{v^{2}}{2} = \frac{q}{m} [a x - \frac{b x^{2}}{2}] \Rightarrow v = \sqrt{\frac{2 q}{m}} {[a x - \frac{b x^{2}}{2}]}^{\frac{1}{2}}$

Now the particle comes to rest for the first time when v=0

$v = 0 \Rightarrow x = \frac{2 a}{b}$

Hence acceleration at that time

$a = \frac{q}{m} (a - b . \frac{2 a}{b}) = - \frac{q a}{m}$

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