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

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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 $y = h/2000$

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(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 $A$.

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