SATHEE: Electrostatics Question 717

Electrostatics Question 717

Question: A particle of charge q and mass m moves rectilinearly under the action of electric field $E = A - B x,$ 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) $\frac{2 A}{B}, 0$

B) $0, - \frac{q A}{m}$

C) $\frac{2 A}{B}, - \frac{q A}{m}$

D) $\frac{- 2 A}{B}, - \frac{q A}{m}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $F = q E = q (A - B x)$

$m a = q (A - B x) \Rightarrow a = \frac{q}{m} (A - B x)$ .(i)

$\frac{v d v}{d x} = \frac{q}{m} (A - B x); v d v = \frac{q}{m} q (A - B x) d x$

$\int_{0}^{0} v d v = \frac{q}{m} \int_{0}^{x} (A - B x) d x; A x - \frac{B x^{2}}{2} = 0$

$x = 0, x = \frac{2 A}{B}$ -(ii)

From eqs. (i) and (ii) $\frac{q}{m} (A - B x) = \frac{q}{m} (A - B \times \frac{2 A}{B}) = \frac{- a A}{m}$

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Electrostatics Question 717

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