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(B) The force on a charged particle in a uniform electric field $E$ is given by $F = qE$. According to Newton's second law,$F = ma$,so the acceleratio

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$(m_p/m_e)^{1/2}$

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(B) The force on a charged particle in a uniform electric field $E$ is given by $F = qE$. According to Newton's second law,$F = ma$,so the acceleration is $a = qE/m$.Using the equation of motion $s = ut + (1/2)at^2$,and since the particles start from rest $(u = 0)$,the distance $s$ covered in time $t$ is $s = (1/2)(qE/m)t^2$.For the electron: $s = (1/2)(eE/m_e)t_1^2$.For the proton: $s = (1/2)(eE/m_p)t_2^2$.Since the distance $s$ is the same for both,we equate the two expressions:$(1/2)(eE/m_e)t_1^2 = (1/2)(eE/m_p)t_2^2$.Simplifying this,we get $t_1^2/m_e = t_2^2/m_p$.Therefore,$t_2^2/t_1^2 = m_p/m_e$.Taking the square root of both sides,we find the ratio $t_2/t_1 = (m_p/m_e)^{1/2}$.

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