JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 22)
A particle of charge q and mass m is subjected to an electric field
E = E0 (1 – $$a$$x2) in the x-direction, where $$a$$ and E0 are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :
E = E0 (1 – $$a$$x2) in the x-direction, where $$a$$ and E0 are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :
$$a$$
$$\sqrt {{2 \over a}} $$
$$\sqrt {{3 \over a}} $$
$$\sqrt {{1 \over a}} $$
Explanation
$$W = \Delta KE$$
As inital and final kinetic energy both are zero so $$\Delta KE$$ = 0
$$ \therefore $$ W = 0
$$ \Rightarrow $$ $$\int\limits_0^x {Fdx} = 0$$
$$ \Rightarrow $$ $$q\int\limits_0^x {{E_0}\left( {1 - a{x^2}} \right)dx} = 0$$
$$ \Rightarrow $$ $$q{E_0}\left[ {\int\limits_0^x {dx - a} \int\limits_0^x {{x^2}dx} } \right] = 0$$
$$ \Rightarrow $$$$q{E_0}\left[ {x - {{a{x^3}} \over 3}} \right] = 0$$
$$ \Rightarrow $$ $$x\left( {1 - {{a{x^2}} \over 3}} \right) = 0$$
$$ \Rightarrow $$ $$x = 0,$$ $${1 - {{a{x^2}} \over 3}}$$ = 0
$$ \Rightarrow $$ $${{{a{x^2}} \over 3}}$$ = 1
$$ \Rightarrow $$ $${x = \sqrt {{3 \over a}} }$$
As inital and final kinetic energy both are zero so $$\Delta KE$$ = 0
$$ \therefore $$ W = 0
$$ \Rightarrow $$ $$\int\limits_0^x {Fdx} = 0$$
$$ \Rightarrow $$ $$q\int\limits_0^x {{E_0}\left( {1 - a{x^2}} \right)dx} = 0$$
$$ \Rightarrow $$ $$q{E_0}\left[ {\int\limits_0^x {dx - a} \int\limits_0^x {{x^2}dx} } \right] = 0$$
$$ \Rightarrow $$$$q{E_0}\left[ {x - {{a{x^3}} \over 3}} \right] = 0$$
$$ \Rightarrow $$ $$x\left( {1 - {{a{x^2}} \over 3}} \right) = 0$$
$$ \Rightarrow $$ $$x = 0,$$ $${1 - {{a{x^2}} \over 3}}$$ = 0
$$ \Rightarrow $$ $${{{a{x^2}} \over 3}}$$ = 1
$$ \Rightarrow $$ $${x = \sqrt {{3 \over a}} }$$
Comments (0)
