A proton is located at coordinates (x, y) = ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The proton has charge $+e$ and mass $m_p$. Its acceleration in the positive $y$ direction is $a_p = eE/m_p$. Its position at time $t$ is $y_p(t) =

Vedclass · Accepted Answer

at about $y = h/2000$

Vedclass · Answer

(A) The proton has charge $+e$ and mass $m_p$. Its acceleration in the positive $y$ direction is $a_p = eE/m_p$. Its position at time $t$ is $y_p(t) = \frac{1}{2} a_p t^2 = \frac{eE}{2m_p} t^2$.The electron has charge $-e$ and mass $m_e$. It starts at $y = h$. Since it has a negative charge,the electric field $E$ exerts a force in the negative $y$ direction. Its acceleration is $a_e = eE/m_e$ downwards. Its position at time $t$ is $y_e(t) = h - \frac{1}{2} a_e t^2 = h - \frac{eE}{2m_e} t^2$.Setting $y_p(t) = y_e(t)$,we get:$\frac{eE}{2m_p} t^2 = h - \frac{eE}{2m_e} t^2$$\frac{eE}{2} t^2 (\frac{1}{m_p} + \frac{1}{m_e}) = h$We want to find the $y$ coordinate $y_p$ where they meet:$y_p = \frac{eE}{2m_p} t^2$From the equation above,$\frac{eE}{2} t^2 = \frac{h}{(\frac{1}{m_p} + \frac{1}{m_e})} = \frac{h m_p m_e}{m_p + m_e}$.Substituting this into the expression for $y_p$:$y_p = \frac{1}{m_p} \cdot \frac{h m_p m_e}{m_p + m_e} = h \frac{m_e}{m_p + m_e}$.Given $m_p \approx 1836 m_e$,we have:$y_p = h \frac{m_e}{1836 m_e + m_e} = h \frac{1}{1837} \approx h/1837$.Since $1837$ is close to $2000$,the correct option is $A$.

Similar Questions

$A$ particle of mass $m$ and charge $q$ travelling with a velocity $v$ along the $x$-axis enters a uniform electric field $\overrightarrow{E}$ directed along the $y$-axis. What will be the trajectory of the particle?

$A$ particle having mass $m$ and charge $q$ is at rest. On applying a uniform electric field $E$ on it,it starts moving. When this particle travels a distance $x$ in the direction of the force,its kinetic energy will be . . . . . . .

$A$ charged particle is free to move in an electric field. It will travel

Vedclass Test Series

Exam Paper Generator

Online Exam Module