Question

A particle (q m) moves under the action of electric field bar(E)=(A-Bx)hat i where A and B are positive constant.If velocity of particle at x=0 is zero then maximum velocity attained by particle is

Answer 1

Answer:

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Answer 2

The maximum velocity attained by the particle is

$\boxed{A\sqrt{\frac{q}{Bm}}}$

Explanation:

Given

Electric Field

$\vec E=(A-Bx)\hat i$

Charge of the particle is q and mass is m

The force acting on the particles

$\vec F=q\vec E$

$\implies \vec F=q(A-Bx)\hat i$

Therefore, acceleration of the particle

$\vec a=\frac{\vec F}{m}$

$\implies \vec a=\frac{q(A-Bx)}{m}\hat i$

The particle will have velocity in the x direction only

We know that

$a=v\frac{dv}{dx}$

$\implies v\frac{dv}{dx}=\frac{q(A-Bx)}{m}$

$\implies vdv=\frac{q(A-Bx)}{m}dx$

$\implies mvdv=q(A-Bx)dx$

$\implies m\int_0^vvdv=q\int_0^x(A-Bx)dx$

$\implies m\frac{v^2}{2}\Bigr|_0^v=q(Ax-B\frac{x^2}{2})\Bigr|_0^x$

$\implies \frac{mv^2}{2}=q(Ax-\frac{Bx^2}{2})$

$\implies \frac{mv^2}{2}=-q\frac{B}{2}(x^2-\frac{2A}{B}x)$

$\implies \frac{mv^2}{2}=-q\frac{B}{2}(x^2-\frac{2A}{B}x+(\frac{A}{B})^2-(\frac{A}{B})^2)$

$\implies \frac{mv^2}{2}=-q\frac{B}{2}(x-\frac{A}{B})^2+q\frac{B}{2}\frac{A^2}{B^2}$

Maximum velocity will be when

$x=\frac{A}{B}$

The maximum velocity is

$v_{max}^2=\frac{1}{m}(\frac{qA^2}{B})$

$\implies v_{max}=A\sqrt{\frac{q}{Bm}}$

Hope this answer is helpful.

Know More:

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